Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation...
Transcript of Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation...
![Page 1: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/1.jpg)
Two Binomial Coefficient Analogues
Bruce SaganDepartment of Mathematics
Michigan State UniversityEast Lansing, MI 48824-1027
[email protected]/˜sagan
March 4, 2013
![Page 2: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/2.jpg)
The theme
Variation 1: binomial coefficients
Variation 2: q-binomial coefficients
Variation 3: fibonomial coefficients
Coda: an open question and bibliography
![Page 3: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/3.jpg)
Outline
The theme
Variation 1: binomial coefficients
Variation 2: q-binomial coefficients
Variation 3: fibonomial coefficients
Coda: an open question and bibliography
![Page 4: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/4.jpg)
Suppose we are given a sequence
a0, a1, a2, . . .
defined so that we only know that the an are rational numbers.
However, it appears as if the an are actually nonegative integers.How would we prove this? There are two standard techniques.
1. Find a recursion for the an and use induction.
2. Find a combinatorial interpretation for the an. In other words,find sets S0,S1, S2, . . . such that, for all n,
an = #Sn
where # denotes cardinality.
![Page 5: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/5.jpg)
Suppose we are given a sequence
a0, a1, a2, . . .
defined so that we only know that the an are rational numbers.However, it appears as if the an are actually nonegative integers.
How would we prove this? There are two standard techniques.
1. Find a recursion for the an and use induction.
2. Find a combinatorial interpretation for the an. In other words,find sets S0,S1, S2, . . . such that, for all n,
an = #Sn
where # denotes cardinality.
![Page 6: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/6.jpg)
Suppose we are given a sequence
a0, a1, a2, . . .
defined so that we only know that the an are rational numbers.However, it appears as if the an are actually nonegative integers.How would we prove this?
There are two standard techniques.
1. Find a recursion for the an and use induction.
2. Find a combinatorial interpretation for the an. In other words,find sets S0,S1, S2, . . . such that, for all n,
an = #Sn
where # denotes cardinality.
![Page 7: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/7.jpg)
Suppose we are given a sequence
a0, a1, a2, . . .
defined so that we only know that the an are rational numbers.However, it appears as if the an are actually nonegative integers.How would we prove this? There are two standard techniques.
1. Find a recursion for the an and use induction.
2. Find a combinatorial interpretation for the an. In other words,find sets S0,S1, S2, . . . such that, for all n,
an = #Sn
where # denotes cardinality.
![Page 8: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/8.jpg)
Suppose we are given a sequence
a0, a1, a2, . . .
defined so that we only know that the an are rational numbers.However, it appears as if the an are actually nonegative integers.How would we prove this? There are two standard techniques.
1. Find a recursion for the an and use induction.
2. Find a combinatorial interpretation for the an. In other words,find sets S0,S1, S2, . . . such that, for all n,
an = #Sn
where # denotes cardinality.
![Page 9: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/9.jpg)
Suppose we are given a sequence
a0, a1, a2, . . .
defined so that we only know that the an are rational numbers.However, it appears as if the an are actually nonegative integers.How would we prove this? There are two standard techniques.
1. Find a recursion for the an and use induction.
2. Find a combinatorial interpretation for the an.
In other words,find sets S0,S1, S2, . . . such that, for all n,
an = #Sn
where # denotes cardinality.
![Page 10: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/10.jpg)
Suppose we are given a sequence
a0, a1, a2, . . .
defined so that we only know that the an are rational numbers.However, it appears as if the an are actually nonegative integers.How would we prove this? There are two standard techniques.
1. Find a recursion for the an and use induction.
2. Find a combinatorial interpretation for the an. In other words,find sets S0, S1, S2, . . . such that, for all n,
an = #Sn
where # denotes cardinality.
![Page 11: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/11.jpg)
Outline
The theme
Variation 1: binomial coefficients
Variation 2: q-binomial coefficients
Variation 3: fibonomial coefficients
Coda: an open question and bibliography
![Page 12: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/12.jpg)
Let N = {0, 1, 2, . . .}.
Suppose n, k ∈ N with 0 ≤ k ≤ n. Definethe corresponding binomial coefficient by(
n
k
)=
n!
k!(n − k)!.
Ex. We have (4
2
)=
4!
2!2!=
4 · 32 · 1
= 6.
Note that it is not clear from this definition that(nk
)is an integer.
Proposition
The binomial coefficients satisfy(n0
)=(nn
)= 1 and, for 0 < k < n,(
n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Since the sum of two integers is an integer, induction on n gives:
Corollary
For all 0 ≤ k ≤ n we have(nk
)∈ N.
![Page 13: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/13.jpg)
Let N = {0, 1, 2, . . .}. Suppose n, k ∈ N with 0 ≤ k ≤ n.
Definethe corresponding binomial coefficient by(
n
k
)=
n!
k!(n − k)!.
Ex. We have (4
2
)=
4!
2!2!=
4 · 32 · 1
= 6.
Note that it is not clear from this definition that(nk
)is an integer.
Proposition
The binomial coefficients satisfy(n0
)=(nn
)= 1 and, for 0 < k < n,(
n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Since the sum of two integers is an integer, induction on n gives:
Corollary
For all 0 ≤ k ≤ n we have(nk
)∈ N.
![Page 14: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/14.jpg)
Let N = {0, 1, 2, . . .}. Suppose n, k ∈ N with 0 ≤ k ≤ n. Definethe corresponding binomial coefficient by(
n
k
)=
n!
k!(n − k)!.
Ex. We have (4
2
)=
4!
2!2!=
4 · 32 · 1
= 6.
Note that it is not clear from this definition that(nk
)is an integer.
Proposition
The binomial coefficients satisfy(n0
)=(nn
)= 1 and, for 0 < k < n,(
n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Since the sum of two integers is an integer, induction on n gives:
Corollary
For all 0 ≤ k ≤ n we have(nk
)∈ N.
![Page 15: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/15.jpg)
Let N = {0, 1, 2, . . .}. Suppose n, k ∈ N with 0 ≤ k ≤ n. Definethe corresponding binomial coefficient by(
n
k
)=
n!
k!(n − k)!.
Ex. We have (4
2
)=
4!
2!2!=
4 · 32 · 1
= 6.
Note that it is not clear from this definition that(nk
)is an integer.
Proposition
The binomial coefficients satisfy(n0
)=(nn
)= 1 and, for 0 < k < n,(
n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Since the sum of two integers is an integer, induction on n gives:
Corollary
For all 0 ≤ k ≤ n we have(nk
)∈ N.
![Page 16: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/16.jpg)
Let N = {0, 1, 2, . . .}. Suppose n, k ∈ N with 0 ≤ k ≤ n. Definethe corresponding binomial coefficient by(
n
k
)=
n!
k!(n − k)!.
Ex. We have (4
2
)=
4!
2!2!=
4 · 32 · 1
= 6.
Note that it is not clear from this definition that(nk
)is an integer.
Proposition
The binomial coefficients satisfy(n0
)=(nn
)= 1 and, for 0 < k < n,(
n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Since the sum of two integers is an integer, induction on n gives:
Corollary
For all 0 ≤ k ≤ n we have(nk
)∈ N.
![Page 17: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/17.jpg)
Let N = {0, 1, 2, . . .}. Suppose n, k ∈ N with 0 ≤ k ≤ n. Definethe corresponding binomial coefficient by(
n
k
)=
n!
k!(n − k)!.
Ex. We have (4
2
)=
4!
2!2!=
4 · 32 · 1
= 6.
Note that it is not clear from this definition that(nk
)is an integer.
Proposition
The binomial coefficients satisfy(n0
)=(nn
)= 1 and, for 0 < k < n,(
n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Since the sum of two integers is an integer, induction on n gives:
Corollary
For all 0 ≤ k ≤ n we have(nk
)∈ N.
![Page 18: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/18.jpg)
Let N = {0, 1, 2, . . .}. Suppose n, k ∈ N with 0 ≤ k ≤ n. Definethe corresponding binomial coefficient by(
n
k
)=
n!
k!(n − k)!.
Ex. We have (4
2
)=
4!
2!2!=
4 · 32 · 1
= 6.
Note that it is not clear from this definition that(nk
)is an integer.
Proposition
The binomial coefficients satisfy(n0
)=(nn
)= 1 and, for 0 < k < n,(
n
k
)=
(n − 1
k
)+
(n − 1
k − 1
).
Since the sum of two integers is an integer, induction on n gives:
Corollary
For all 0 ≤ k ≤ n we have(nk
)∈ N.
![Page 19: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/19.jpg)
(a) Subsets.
Let
Sn,k = {S : S is a k-element subset of {1, . . . , n}}.
Ex. We have
S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.
Note
#S4,2 = 6 =
(4
2
).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Sn,k .
![Page 20: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/20.jpg)
(a) Subsets.Let
Sn,k = {S : S is a k-element subset of {1, . . . , n}}.
Ex. We have
S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.
Note
#S4,2 = 6 =
(4
2
).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Sn,k .
![Page 21: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/21.jpg)
(a) Subsets.Let
Sn,k = {S : S is a k-element subset of {1, . . . , n}}.
Ex. We have
S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.
Note
#S4,2 = 6 =
(4
2
).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Sn,k .
![Page 22: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/22.jpg)
(a) Subsets.Let
Sn,k = {S : S is a k-element subset of {1, . . . , n}}.
Ex. We have
S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.
Note
#S4,2 = 6 =
(4
2
).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Sn,k .
![Page 23: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/23.jpg)
(a) Subsets.Let
Sn,k = {S : S is a k-element subset of {1, . . . , n}}.
Ex. We have
S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.
Note
#S4,2 = 6 =
(4
2
).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Sn,k .
![Page 24: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/24.jpg)
(b) Words.
A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let
Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.
Ex. We have
W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.
Note that
#W4,2 = 6 =
(4
2
).
Any w = a1 . . . an ∈ Wn,k can be obtained by choosing k of the npositions to be zeros (and the rest will be ones by default).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Wn,k .
![Page 25: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/25.jpg)
(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all i
Ex. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let
Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.
Ex. We have
W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.
Note that
#W4,2 = 6 =
(4
2
).
Any w = a1 . . . an ∈ Wn,k can be obtained by choosing k of the npositions to be zeros (and the rest will be ones by default).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Wn,k .
![Page 26: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/26.jpg)
(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.
Let
Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.
Ex. We have
W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.
Note that
#W4,2 = 6 =
(4
2
).
Any w = a1 . . . an ∈ Wn,k can be obtained by choosing k of the npositions to be zeros (and the rest will be ones by default).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Wn,k .
![Page 27: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/27.jpg)
(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let
Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.
Ex. We have
W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.
Note that
#W4,2 = 6 =
(4
2
).
Any w = a1 . . . an ∈ Wn,k can be obtained by choosing k of the npositions to be zeros (and the rest will be ones by default).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Wn,k .
![Page 28: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/28.jpg)
(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let
Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.
Ex. We have
W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.
Note that
#W4,2 = 6 =
(4
2
).
Any w = a1 . . . an ∈ Wn,k can be obtained by choosing k of the npositions to be zeros (and the rest will be ones by default).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Wn,k .
![Page 29: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/29.jpg)
(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let
Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.
Ex. We have
W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.
Note that
#W4,2 = 6 =
(4
2
).
Any w = a1 . . . an ∈ Wn,k can be obtained by choosing k of the npositions to be zeros (and the rest will be ones by default).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Wn,k .
![Page 30: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/30.jpg)
(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let
Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.
Ex. We have
W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.
Note that
#W4,2 = 6 =
(4
2
).
Any w = a1 . . . an ∈ Wn,k can be obtained by choosing k of the npositions to be zeros (and the rest will be ones by default).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Wn,k .
![Page 31: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/31.jpg)
(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let
Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.
Ex. We have
W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.
Note that
#W4,2 = 6 =
(4
2
).
Any w = a1 . . . an ∈ Wn,k can be obtained by choosing k of the npositions to be zeros (and the rest will be ones by default).
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Wn,k .
![Page 32: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/32.jpg)
(c) Lattice paths.
A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).Ex. We have
P = ENEEN =
Let
Pn,k = {P = s1 . . . sn : P has k N-steps and n − k E -steps}
There is a bijection f : Pn,k →Wn,k where w = f (P) is obtainedby replacing each N by a 0 and each E by a 1.Ex. We have
P = ENEENf←→ w = 10110 =
1
01 1 0
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Pn,k .
![Page 33: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/33.jpg)
(c) Lattice paths.A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).
Ex. We have
P = ENEEN =
Let
Pn,k = {P = s1 . . . sn : P has k N-steps and n − k E -steps}
There is a bijection f : Pn,k →Wn,k where w = f (P) is obtainedby replacing each N by a 0 and each E by a 1.Ex. We have
P = ENEENf←→ w = 10110 =
1
01 1 0
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Pn,k .
![Page 34: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/34.jpg)
(c) Lattice paths.A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).Ex. We have
P = ENEEN =
Let
Pn,k = {P = s1 . . . sn : P has k N-steps and n − k E -steps}
There is a bijection f : Pn,k →Wn,k where w = f (P) is obtainedby replacing each N by a 0 and each E by a 1.Ex. We have
P = ENEENf←→ w = 10110 =
1
01 1 0
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Pn,k .
![Page 35: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/35.jpg)
(c) Lattice paths.A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).Ex. We have
P = ENEEN =
Let
Pn,k = {P = s1 . . . sn : P has k N-steps and n − k E -steps}
There is a bijection f : Pn,k →Wn,k where w = f (P) is obtainedby replacing each N by a 0 and each E by a 1.Ex. We have
P = ENEENf←→ w = 10110 =
1
01 1 0
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Pn,k .
![Page 36: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/36.jpg)
(c) Lattice paths.A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).Ex. We have
P = ENEEN =
Let
Pn,k = {P = s1 . . . sn : P has k N-steps and n − k E -steps}
There is a bijection f : Pn,k →Wn,k where w = f (P) is obtainedby replacing each N by a 0 and each E by a 1.
Ex. We have
P = ENEENf←→ w = 10110 =
1
01 1 0
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Pn,k .
![Page 37: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/37.jpg)
(c) Lattice paths.A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).Ex. We have
P = ENEEN =
Let
Pn,k = {P = s1 . . . sn : P has k N-steps and n − k E -steps}
There is a bijection f : Pn,k →Wn,k where w = f (P) is obtainedby replacing each N by a 0 and each E by a 1.Ex. We have
P = ENEENf←→ w = 10110 =
1
01 1 0
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Pn,k .
![Page 38: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/38.jpg)
(c) Lattice paths.A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).Ex. We have
P = ENEEN =
Let
Pn,k = {P = s1 . . . sn : P has k N-steps and n − k E -steps}
There is a bijection f : Pn,k →Wn,k where w = f (P) is obtainedby replacing each N by a 0 and each E by a 1.Ex. We have
P = ENEENf←→ w = 10110 =
1
01 1 0
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Pn,k .
![Page 39: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/39.jpg)
(d) Partitions.
An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) = ⊆ 3× 4: g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 40: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/40.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm).
The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) = ⊆ 3× 4: g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 41: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/41.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm).
The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row i
Ex. Suppose λ = (3, 2, 2).
Then
λ = (3, 2, 2) = ⊆ 3× 4: g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 42: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/42.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2).
Then
λ = (3, 2, 2) = ⊆ 3× 4: g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 43: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/43.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) =
⊆ 3× 4: g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 44: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/44.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) =
⊆ 3× 4: g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns.
Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 45: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/45.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) = ⊆ 3× 4:
g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns.
Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 46: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/46.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) = ⊆ 3× 4:
g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns.
Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 47: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/47.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) = ⊆ 3× 4:
g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 48: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/48.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) = ⊆ 3× 4:
g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k :
given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 49: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/49.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) = ⊆ 3× 4:
g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 50: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/50.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) = ⊆ 3× 4: g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 51: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/51.jpg)
(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then
λ = (3, 2, 2) = ⊆ 3× 4: g←→
We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let
Ln,k = {λ : λ ⊆ k × (n − k)}.
There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.
Proposition
For 0 ≤ k ≤ n we have(nk
)= #Ln,k .
![Page 52: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/52.jpg)
Outline
The theme
Variation 1: binomial coefficients
Variation 2: q-binomial coefficients
Variation 3: fibonomial coefficients
Coda: an open question and bibliography
![Page 53: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/53.jpg)
A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O.
The standardq-analogue of n ∈ N is the polynomial
[n] = 1 + q + q2 + · · ·+ qn−1.
Ex. We have [4] = 1 + q + q2 + q3.Note that
[n]|q=1 =
n︷ ︸︸ ︷1 + 1 + · · ·+ 1 = n.
A q-factorial is [n]! = [1][2] · · · [n]. For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n
k
]=
[n]!
[k]![n − k]!.
Ex. We have[4
2
]=
[4]!
[2]![2]!=
[4][3]
[2][1]= 1 + q + 2q2 + q3 + q4.
Note that it is not clear from the definition that[ nk
]is always in
N[q], the set of polynomials in q with coefficients in N.
![Page 54: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/54.jpg)
A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O. The standardq-analogue of n ∈ N is the polynomial
[n] = 1 + q + q2 + · · ·+ qn−1.
Ex. We have [4] = 1 + q + q2 + q3.Note that
[n]|q=1 =
n︷ ︸︸ ︷1 + 1 + · · ·+ 1 = n.
A q-factorial is [n]! = [1][2] · · · [n]. For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n
k
]=
[n]!
[k]![n − k]!.
Ex. We have[4
2
]=
[4]!
[2]![2]!=
[4][3]
[2][1]= 1 + q + 2q2 + q3 + q4.
Note that it is not clear from the definition that[ nk
]is always in
N[q], the set of polynomials in q with coefficients in N.
![Page 55: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/55.jpg)
A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O. The standardq-analogue of n ∈ N is the polynomial
[n] = 1 + q + q2 + · · ·+ qn−1.
Ex. We have [4] = 1 + q + q2 + q3.
Note that
[n]|q=1 =
n︷ ︸︸ ︷1 + 1 + · · ·+ 1 = n.
A q-factorial is [n]! = [1][2] · · · [n]. For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n
k
]=
[n]!
[k]![n − k]!.
Ex. We have[4
2
]=
[4]!
[2]![2]!=
[4][3]
[2][1]= 1 + q + 2q2 + q3 + q4.
Note that it is not clear from the definition that[ nk
]is always in
N[q], the set of polynomials in q with coefficients in N.
![Page 56: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/56.jpg)
A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O. The standardq-analogue of n ∈ N is the polynomial
[n] = 1 + q + q2 + · · ·+ qn−1.
Ex. We have [4] = 1 + q + q2 + q3.Note that
[n]|q=1 =
n︷ ︸︸ ︷1 + 1 + · · ·+ 1 = n.
A q-factorial is [n]! = [1][2] · · · [n]. For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n
k
]=
[n]!
[k]![n − k]!.
Ex. We have[4
2
]=
[4]!
[2]![2]!=
[4][3]
[2][1]= 1 + q + 2q2 + q3 + q4.
Note that it is not clear from the definition that[ nk
]is always in
N[q], the set of polynomials in q with coefficients in N.
![Page 57: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/57.jpg)
A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O. The standardq-analogue of n ∈ N is the polynomial
[n] = 1 + q + q2 + · · ·+ qn−1.
Ex. We have [4] = 1 + q + q2 + q3.Note that
[n]|q=1 =
n︷ ︸︸ ︷1 + 1 + · · ·+ 1 = n.
A q-factorial is [n]! = [1][2] · · · [n].
For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n
k
]=
[n]!
[k]![n − k]!.
Ex. We have[4
2
]=
[4]!
[2]![2]!=
[4][3]
[2][1]= 1 + q + 2q2 + q3 + q4.
Note that it is not clear from the definition that[ nk
]is always in
N[q], the set of polynomials in q with coefficients in N.
![Page 58: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/58.jpg)
A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O. The standardq-analogue of n ∈ N is the polynomial
[n] = 1 + q + q2 + · · ·+ qn−1.
Ex. We have [4] = 1 + q + q2 + q3.Note that
[n]|q=1 =
n︷ ︸︸ ︷1 + 1 + · · ·+ 1 = n.
A q-factorial is [n]! = [1][2] · · · [n]. For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n
k
]=
[n]!
[k]![n − k]!.
Ex. We have[4
2
]=
[4]!
[2]![2]!=
[4][3]
[2][1]= 1 + q + 2q2 + q3 + q4.
Note that it is not clear from the definition that[ nk
]is always in
N[q], the set of polynomials in q with coefficients in N.
![Page 59: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/59.jpg)
A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O. The standardq-analogue of n ∈ N is the polynomial
[n] = 1 + q + q2 + · · ·+ qn−1.
Ex. We have [4] = 1 + q + q2 + q3.Note that
[n]|q=1 =
n︷ ︸︸ ︷1 + 1 + · · ·+ 1 = n.
A q-factorial is [n]! = [1][2] · · · [n]. For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n
k
]=
[n]!
[k]![n − k]!.
Ex. We have[4
2
]=
[4]!
[2]![2]!=
[4][3]
[2][1]= 1 + q + 2q2 + q3 + q4.
Note that it is not clear from the definition that[ nk
]is always in
N[q], the set of polynomials in q with coefficients in N.
![Page 60: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/60.jpg)
A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O. The standardq-analogue of n ∈ N is the polynomial
[n] = 1 + q + q2 + · · ·+ qn−1.
Ex. We have [4] = 1 + q + q2 + q3.Note that
[n]|q=1 =
n︷ ︸︸ ︷1 + 1 + · · ·+ 1 = n.
A q-factorial is [n]! = [1][2] · · · [n]. For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n
k
]=
[n]!
[k]![n − k]!.
Ex. We have[4
2
]=
[4]!
[2]![2]!=
[4][3]
[2][1]= 1 + q + 2q2 + q3 + q4.
Note that it is not clear from the definition that[ nk
]is always in
N[q], the set of polynomials in q with coefficients in N.
![Page 61: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/61.jpg)
Here is a q-analogue for the boundary conditions and recurrencerelation for the binomial coefficients.
TheoremThe q-binomial coefficients satisfy
[n0
]=[nn
]= 1 and, for
0 < k < n, [n
k
]= qk
[n − 1
k
]+
[n − 1
k − 1
]
=
[n − 1
k
]+ qn−k
[n − 1
k − 1
].
Since sums and products of elements of N[q] are again in N[q], weimmediately get the following result.
Corollary
For all 0 ≤ k ≤ n we have[ nk
]∈ N[q].
![Page 62: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/62.jpg)
Here is a q-analogue for the boundary conditions and recurrencerelation for the binomial coefficients.
TheoremThe q-binomial coefficients satisfy
[n0
]=[nn
]= 1 and, for
0 < k < n, [n
k
]= qk
[n − 1
k
]+
[n − 1
k − 1
]
=
[n − 1
k
]+ qn−k
[n − 1
k − 1
].
Since sums and products of elements of N[q] are again in N[q], weimmediately get the following result.
Corollary
For all 0 ≤ k ≤ n we have[ nk
]∈ N[q].
![Page 63: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/63.jpg)
Here is a q-analogue for the boundary conditions and recurrencerelation for the binomial coefficients.
TheoremThe q-binomial coefficients satisfy
[n0
]=[nn
]= 1 and, for
0 < k < n, [n
k
]= qk
[n − 1
k
]+
[n − 1
k − 1
]
=
[n − 1
k
]+ qn−k
[n − 1
k − 1
].
Since sums and products of elements of N[q] are again in N[q], weimmediately get the following result.
Corollary
For all 0 ≤ k ≤ n we have[ nk
]∈ N[q].
![Page 64: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/64.jpg)
Here is a q-analogue for the boundary conditions and recurrencerelation for the binomial coefficients.
TheoremThe q-binomial coefficients satisfy
[n0
]=[nn
]= 1 and, for
0 < k < n, [n
k
]= qk
[n − 1
k
]+
[n − 1
k − 1
]
=
[n − 1
k
]+ qn−k
[n − 1
k − 1
].
Since sums and products of elements of N[q] are again in N[q], weimmediately get the following result.
Corollary
For all 0 ≤ k ≤ n we have[ nk
]∈ N[q].
![Page 65: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/65.jpg)
(a) Words.
If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 66: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/66.jpg)
(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 67: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/67.jpg)
(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)}
and invw = 4.Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 68: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/68.jpg)
(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)}
and invw = 4.Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 69: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/69.jpg)
(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.
Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 70: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/70.jpg)
(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 71: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/71.jpg)
(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100
I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 72: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/72.jpg)
(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 73: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/73.jpg)
(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 74: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/74.jpg)
(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is
Invw = {(i , j) : i < j and ai > aj}.
The corresponding inversion number is
invw = # Invw .
Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.Consider the inversion generating function
In,k(q) =∑
w∈Wn,k
qinvw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= In,k(q).
![Page 75: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/75.jpg)
(b) Even more words.
If w = a1 . . . an is a word over N then the major index of w is
majw =∑
ai>ai+1
i .
Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.Consider the major index generating function
Mn,k(q) =∑
w∈Wn,k
qmajw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Mn,k(q).
![Page 76: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/76.jpg)
(b) Even more words.If w = a1 . . . an is a word over N then the major index of w is
majw =∑
ai>ai+1
i .
Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.Consider the major index generating function
Mn,k(q) =∑
w∈Wn,k
qmajw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Mn,k(q).
![Page 77: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/77.jpg)
(b) Even more words.If w = a1 . . . an is a word over N then the major index of w is
majw =∑
ai>ai+1
i .
Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.
Consider the major index generating function
Mn,k(q) =∑
w∈Wn,k
qmajw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Mn,k(q).
![Page 78: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/78.jpg)
(b) Even more words.If w = a1 . . . an is a word over N then the major index of w is
majw =∑
ai>ai+1
i .
Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.Consider the major index generating function
Mn,k(q) =∑
w∈Wn,k
qmajw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Mn,k(q).
![Page 79: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/79.jpg)
(b) Even more words.If w = a1 . . . an is a word over N then the major index of w is
majw =∑
ai>ai+1
i .
Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.Consider the major index generating function
Mn,k(q) =∑
w∈Wn,k
qmajw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100
M4,2(q) = q0 + q2 + q3 + q + q4 + q2
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Mn,k(q).
![Page 80: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/80.jpg)
(b) Even more words.If w = a1 . . . an is a word over N then the major index of w is
majw =∑
ai>ai+1
i .
Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.Consider the major index generating function
Mn,k(q) =∑
w∈Wn,k
qmajw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Mn,k(q).
![Page 81: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/81.jpg)
(b) Even more words.If w = a1 . . . an is a word over N then the major index of w is
majw =∑
ai>ai+1
i .
Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.Consider the major index generating function
Mn,k(q) =∑
w∈Wn,k
qmajw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Mn,k(q).
![Page 82: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/82.jpg)
(b) Even more words.If w = a1 . . . an is a word over N then the major index of w is
majw =∑
ai>ai+1
i .
Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.Consider the major index generating function
Mn,k(q) =∑
w∈Wn,k
qmajw .
Ex. When n = 4 and k = 2,
W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2
=[42
].
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Mn,k(q).
![Page 83: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/83.jpg)
(c) Partitions.
If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 84: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/84.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 85: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/85.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.
Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 86: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/86.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.
Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 87: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/87.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 88: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/88.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .
In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 89: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/89.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .
Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 90: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/90.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 91: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/91.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 92: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/92.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 1
10
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 93: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/93.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 1
10
10 1
10
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 94: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/94.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 1
10
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 95: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/95.jpg)
(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is
|λ| = λ1 + · · ·+ λm.
Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function
Sn,k(q) =∑λ∈Ln,k
q|λ|.
Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2
w = 10110h←→ λ =
10 1 1
0
1
010 11
0
10 11
0
10 1
10 1 1
0
TheoremFor all 0 ≤ k ≤ n we have
[ nk
]= Sn,k(q).
![Page 96: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/96.jpg)
(d) Subspaces.
Let q be a prime power and Fq be the Galois field with q elements.Consider the n-dimensional vector space Fn
q. Let
Vn,k(q) = {W : W is a k-dimensional subspace of Fnq}.
Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗
] [1 ∗ 0 ∗0 0 1 ∗
] [1 ∗ ∗ 00 0 0 1
][
0 1 0 ∗0 0 1 ∗
] [0 1 ∗ 00 0 0 1
] [0 0 1 00 0 0 1
]where the stars are arbitrary elements of F3. Therefore
#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =
[4
2
]∣∣∣∣q=3
.
Theorem (Knuth, 1971)
For all 0 ≤ k ≤ n and q a prime power we have[ nk
]= #Vn,k(q).
![Page 97: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/97.jpg)
(d) Subspaces.Let q be a prime power and Fq be the Galois field with q elements.
Consider the n-dimensional vector space Fnq. Let
Vn,k(q) = {W : W is a k-dimensional subspace of Fnq}.
Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗
] [1 ∗ 0 ∗0 0 1 ∗
] [1 ∗ ∗ 00 0 0 1
][
0 1 0 ∗0 0 1 ∗
] [0 1 ∗ 00 0 0 1
] [0 0 1 00 0 0 1
]where the stars are arbitrary elements of F3. Therefore
#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =
[4
2
]∣∣∣∣q=3
.
Theorem (Knuth, 1971)
For all 0 ≤ k ≤ n and q a prime power we have[ nk
]= #Vn,k(q).
![Page 98: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/98.jpg)
(d) Subspaces.Let q be a prime power and Fq be the Galois field with q elements.Consider the n-dimensional vector space Fn
q.
Let
Vn,k(q) = {W : W is a k-dimensional subspace of Fnq}.
Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗
] [1 ∗ 0 ∗0 0 1 ∗
] [1 ∗ ∗ 00 0 0 1
][
0 1 0 ∗0 0 1 ∗
] [0 1 ∗ 00 0 0 1
] [0 0 1 00 0 0 1
]where the stars are arbitrary elements of F3. Therefore
#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =
[4
2
]∣∣∣∣q=3
.
Theorem (Knuth, 1971)
For all 0 ≤ k ≤ n and q a prime power we have[ nk
]= #Vn,k(q).
![Page 99: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/99.jpg)
(d) Subspaces.Let q be a prime power and Fq be the Galois field with q elements.Consider the n-dimensional vector space Fn
q. Let
Vn,k(q) = {W : W is a k-dimensional subspace of Fnq}.
Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗
] [1 ∗ 0 ∗0 0 1 ∗
] [1 ∗ ∗ 00 0 0 1
][
0 1 0 ∗0 0 1 ∗
] [0 1 ∗ 00 0 0 1
] [0 0 1 00 0 0 1
]where the stars are arbitrary elements of F3. Therefore
#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =
[4
2
]∣∣∣∣q=3
.
Theorem (Knuth, 1971)
For all 0 ≤ k ≤ n and q a prime power we have[ nk
]= #Vn,k(q).
![Page 100: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/100.jpg)
(d) Subspaces.Let q be a prime power and Fq be the Galois field with q elements.Consider the n-dimensional vector space Fn
q. Let
Vn,k(q) = {W : W is a k-dimensional subspace of Fnq}.
Ex. Let q = 3.
The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗
] [1 ∗ 0 ∗0 0 1 ∗
] [1 ∗ ∗ 00 0 0 1
][
0 1 0 ∗0 0 1 ∗
] [0 1 ∗ 00 0 0 1
] [0 0 1 00 0 0 1
]where the stars are arbitrary elements of F3. Therefore
#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =
[4
2
]∣∣∣∣q=3
.
Theorem (Knuth, 1971)
For all 0 ≤ k ≤ n and q a prime power we have[ nk
]= #Vn,k(q).
![Page 101: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/101.jpg)
(d) Subspaces.Let q be a prime power and Fq be the Galois field with q elements.Consider the n-dimensional vector space Fn
q. Let
Vn,k(q) = {W : W is a k-dimensional subspace of Fnq}.
Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗
] [1 ∗ 0 ∗0 0 1 ∗
] [1 ∗ ∗ 00 0 0 1
][
0 1 0 ∗0 0 1 ∗
] [0 1 ∗ 00 0 0 1
] [0 0 1 00 0 0 1
]where the stars are arbitrary elements of F3.
Therefore
#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =
[4
2
]∣∣∣∣q=3
.
Theorem (Knuth, 1971)
For all 0 ≤ k ≤ n and q a prime power we have[ nk
]= #Vn,k(q).
![Page 102: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/102.jpg)
(d) Subspaces.Let q be a prime power and Fq be the Galois field with q elements.Consider the n-dimensional vector space Fn
q. Let
Vn,k(q) = {W : W is a k-dimensional subspace of Fnq}.
Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗
] [1 ∗ 0 ∗0 0 1 ∗
] [1 ∗ ∗ 00 0 0 1
][
0 1 0 ∗0 0 1 ∗
] [0 1 ∗ 00 0 0 1
] [0 0 1 00 0 0 1
]where the stars are arbitrary elements of F3. Therefore
#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =
[4
2
]∣∣∣∣q=3
.
Theorem (Knuth, 1971)
For all 0 ≤ k ≤ n and q a prime power we have[ nk
]= #Vn,k(q).
![Page 103: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/103.jpg)
(d) Subspaces.Let q be a prime power and Fq be the Galois field with q elements.Consider the n-dimensional vector space Fn
q. Let
Vn,k(q) = {W : W is a k-dimensional subspace of Fnq}.
Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗
] [1 ∗ 0 ∗0 0 1 ∗
] [1 ∗ ∗ 00 0 0 1
][
0 1 0 ∗0 0 1 ∗
] [0 1 ∗ 00 0 0 1
] [0 0 1 00 0 0 1
]where the stars are arbitrary elements of F3. Therefore
#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =
[4
2
]∣∣∣∣q=3
.
Theorem (Knuth, 1971)
For all 0 ≤ k ≤ n and q a prime power we have[ nk
]= #Vn,k(q).
![Page 104: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/104.jpg)
Outline
The theme
Variation 1: binomial coefficients
Variation 2: q-binomial coefficients
Variation 3: fibonomial coefficients
Coda: an open question and bibliography
![Page 105: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/105.jpg)
The Fibonacci numbers are defined by F1 = F2 = 1 and, for n ≥ 2,
Fn = Fn−1 + Fn−2.
Ex. F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, . . .A fibotorial is F !
n = F1F2 · · ·Fn. For 0 ≤ k ≤ n, the correspondingfibonomial is (
n
k
)F
=F !n
F !kF !
n−k.
Ex. We have(6
3
)F
=F !6
F !3F !
3
=F6F5F4
F3F2F1=
8 · 5 · 32 · 1 · 1
= 60.
TheoremThe fibonomials satisfy
(n0
)F
=(nn
)F
= 1 and, for 0 < k < n,(n
k
)F
= Fk+1
(n − 1
k
)F
+ Fn−k−1
(n − 1
k − 1
)F
.
Corollary
For all 0 ≤ k ≤ n we have(nk
)F∈ N.
![Page 106: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/106.jpg)
The Fibonacci numbers are defined by F1 = F2 = 1 and, for n ≥ 2,
Fn = Fn−1 + Fn−2.
Ex. F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, . . .
A fibotorial is F !n = F1F2 · · ·Fn. For 0 ≤ k ≤ n, the corresponding
fibonomial is (n
k
)F
=F !n
F !kF !
n−k.
Ex. We have(6
3
)F
=F !6
F !3F !
3
=F6F5F4
F3F2F1=
8 · 5 · 32 · 1 · 1
= 60.
TheoremThe fibonomials satisfy
(n0
)F
=(nn
)F
= 1 and, for 0 < k < n,(n
k
)F
= Fk+1
(n − 1
k
)F
+ Fn−k−1
(n − 1
k − 1
)F
.
Corollary
For all 0 ≤ k ≤ n we have(nk
)F∈ N.
![Page 107: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/107.jpg)
The Fibonacci numbers are defined by F1 = F2 = 1 and, for n ≥ 2,
Fn = Fn−1 + Fn−2.
Ex. F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, . . .A fibotorial is F !
n = F1F2 · · ·Fn.
For 0 ≤ k ≤ n, the correspondingfibonomial is (
n
k
)F
=F !n
F !kF !
n−k.
Ex. We have(6
3
)F
=F !6
F !3F !
3
=F6F5F4
F3F2F1=
8 · 5 · 32 · 1 · 1
= 60.
TheoremThe fibonomials satisfy
(n0
)F
=(nn
)F
= 1 and, for 0 < k < n,(n
k
)F
= Fk+1
(n − 1
k
)F
+ Fn−k−1
(n − 1
k − 1
)F
.
Corollary
For all 0 ≤ k ≤ n we have(nk
)F∈ N.
![Page 108: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/108.jpg)
The Fibonacci numbers are defined by F1 = F2 = 1 and, for n ≥ 2,
Fn = Fn−1 + Fn−2.
Ex. F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, . . .A fibotorial is F !
n = F1F2 · · ·Fn. For 0 ≤ k ≤ n, the correspondingfibonomial is (
n
k
)F
=F !n
F !kF !
n−k.
Ex. We have(6
3
)F
=F !6
F !3F !
3
=F6F5F4
F3F2F1=
8 · 5 · 32 · 1 · 1
= 60.
TheoremThe fibonomials satisfy
(n0
)F
=(nn
)F
= 1 and, for 0 < k < n,(n
k
)F
= Fk+1
(n − 1
k
)F
+ Fn−k−1
(n − 1
k − 1
)F
.
Corollary
For all 0 ≤ k ≤ n we have(nk
)F∈ N.
![Page 109: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/109.jpg)
The Fibonacci numbers are defined by F1 = F2 = 1 and, for n ≥ 2,
Fn = Fn−1 + Fn−2.
Ex. F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, . . .A fibotorial is F !
n = F1F2 · · ·Fn. For 0 ≤ k ≤ n, the correspondingfibonomial is (
n
k
)F
=F !n
F !kF !
n−k.
Ex. We have(6
3
)F
=F !6
F !3F !
3
=F6F5F4
F3F2F1=
8 · 5 · 32 · 1 · 1
= 60.
TheoremThe fibonomials satisfy
(n0
)F
=(nn
)F
= 1 and, for 0 < k < n,(n
k
)F
= Fk+1
(n − 1
k
)F
+ Fn−k−1
(n − 1
k − 1
)F
.
Corollary
For all 0 ≤ k ≤ n we have(nk
)F∈ N.
![Page 110: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/110.jpg)
The Fibonacci numbers are defined by F1 = F2 = 1 and, for n ≥ 2,
Fn = Fn−1 + Fn−2.
Ex. F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, . . .A fibotorial is F !
n = F1F2 · · ·Fn. For 0 ≤ k ≤ n, the correspondingfibonomial is (
n
k
)F
=F !n
F !kF !
n−k.
Ex. We have(6
3
)F
=F !6
F !3F !
3
=F6F5F4
F3F2F1=
8 · 5 · 32 · 1 · 1
= 60.
TheoremThe fibonomials satisfy
(n0
)F
=(nn
)F
= 1 and, for 0 < k < n,(n
k
)F
= Fk+1
(n − 1
k
)F
+ Fn−k−1
(n − 1
k − 1
)F
.
Corollary
For all 0 ≤ k ≤ n we have(nk
)F∈ N.
![Page 111: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/111.jpg)
The Fibonacci numbers are defined by F1 = F2 = 1 and, for n ≥ 2,
Fn = Fn−1 + Fn−2.
Ex. F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, . . .A fibotorial is F !
n = F1F2 · · ·Fn. For 0 ≤ k ≤ n, the correspondingfibonomial is (
n
k
)F
=F !n
F !kF !
n−k.
Ex. We have(6
3
)F
=F !6
F !3F !
3
=F6F5F4
F3F2F1=
8 · 5 · 32 · 1 · 1
= 60.
TheoremThe fibonomials satisfy
(n0
)F
=(nn
)F
= 1 and, for 0 < k < n,(n
k
)F
= Fk+1
(n − 1
k
)F
+ Fn−k−1
(n − 1
k − 1
)F
.
Corollary
For all 0 ≤ k ≤ n we have(nk
)F∈ N.
![Page 112: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/112.jpg)
S. and Savage were the first to give a simple combinatorialinterpretation of
(nk
)F
.
Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot. Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes. Let
Tn = {T : T a linear tiling of a row of n squares}.
Ex. We have
T3 ={
, ,
}Note that #T3 = 3 = F4.
Proposition
For all n ≥ 1 we have Fn = #Tn−1.
![Page 113: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/113.jpg)
S. and Savage were the first to give a simple combinatorialinterpretation of
(nk
)F
. Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot.
Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes. Let
Tn = {T : T a linear tiling of a row of n squares}.
Ex. We have
T3 ={
, ,
}Note that #T3 = 3 = F4.
Proposition
For all n ≥ 1 we have Fn = #Tn−1.
![Page 114: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/114.jpg)
S. and Savage were the first to give a simple combinatorialinterpretation of
(nk
)F
. Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot. Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes.
Let
Tn = {T : T a linear tiling of a row of n squares}.
Ex. We have
T3 ={
, ,
}Note that #T3 = 3 = F4.
Proposition
For all n ≥ 1 we have Fn = #Tn−1.
![Page 115: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/115.jpg)
S. and Savage were the first to give a simple combinatorialinterpretation of
(nk
)F
. Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot. Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes. Let
Tn = {T : T a linear tiling of a row of n squares}.
Ex. We have
T3 ={
, ,
}Note that #T3 = 3 = F4.
Proposition
For all n ≥ 1 we have Fn = #Tn−1.
![Page 116: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/116.jpg)
S. and Savage were the first to give a simple combinatorialinterpretation of
(nk
)F
. Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot. Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes. Let
Tn = {T : T a linear tiling of a row of n squares}.
Ex. We have
T3 ={
, ,
}
Note that #T3 = 3 = F4.
Proposition
For all n ≥ 1 we have Fn = #Tn−1.
![Page 117: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/117.jpg)
S. and Savage were the first to give a simple combinatorialinterpretation of
(nk
)F
. Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot. Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes. Let
Tn = {T : T a linear tiling of a row of n squares}.
Ex. We have
T3 ={
, ,
}Note that #T3 = 3 = F4.
Proposition
For all n ≥ 1 we have Fn = #Tn−1.
![Page 118: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/118.jpg)
S. and Savage were the first to give a simple combinatorialinterpretation of
(nk
)F
. Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot. Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes. Let
Tn = {T : T a linear tiling of a row of n squares}.
Ex. We have
T3 ={
, ,
}Note that #T3 = 3 = F4.
Proposition
For all n ≥ 1 we have Fn = #Tn−1.
![Page 119: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/119.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi .
Let
Tλ = {T : T is a tiling of λ},Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ. Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 120: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/120.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let
Tλ = {T : T is a tiling of λ},
Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ. Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 121: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/121.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let
Tλ = {T : T is a tiling of λ},
Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2),
∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ. Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 122: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/122.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let
Tλ = {T : T is a tiling of λ},Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2),
∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ. Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 123: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/123.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let
Tλ = {T : T is a tiling of λ},Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ. Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 124: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/124.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let
Tλ = {T : T is a tiling of λ},Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ.
Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 125: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/125.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let
Tλ = {T : T is a tiling of λ},Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ.
Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 126: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/126.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let
Tλ = {T : T is a tiling of λ},Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ. Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 127: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/127.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let
Tλ = {T : T is a tiling of λ},Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ. Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 128: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/128.jpg)
A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let
Tλ = {T : T is a tiling of λ},Dλ = {T ∈ Tλ : every λi begins with a domino}.
Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).
If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where
the λ∗j are the column lengths of (k × l)− λ. Let
Fn,k =⋃
λ⊆k×(n−k)
(Tλ ×Dλ∗) .
Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):
and ∈ F7,3
Proposition (S. and Savage, 2010)
For 0 ≤ k ≤ n we have(nk
)F
= #Fn,k .
![Page 129: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/129.jpg)
Outline
The theme
Variation 1: binomial coefficients
Variation 2: q-binomial coefficients
Variation 3: fibonomial coefficients
Coda: an open question and bibliography
![Page 130: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/130.jpg)
The nth Catalan number is
Cn =1
n + 1
(2n
n
).
Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .
TheoremWe have C0 = 1 and, for n ≥ 1, Cn =
∑n−1i=0 CiCn−i−1.
Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.For example, let
Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.TheoremFor n ≥ 0 we have Cn = #Dn.
Define the fibocatalan numbers to be
Cn,F =1
Fn+1
(2n
n
)F
.
It is not hard to show Cn.F ∈ N for all n. Lou Shapiro asked: Canone find a combinatorial interpretation?
![Page 131: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/131.jpg)
The nth Catalan number is
Cn =1
n + 1
(2n
n
).
Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .
TheoremWe have C0 = 1 and, for n ≥ 1, Cn =
∑n−1i=0 CiCn−i−1.
Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.For example, let
Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.TheoremFor n ≥ 0 we have Cn = #Dn.
Define the fibocatalan numbers to be
Cn,F =1
Fn+1
(2n
n
)F
.
It is not hard to show Cn.F ∈ N for all n. Lou Shapiro asked: Canone find a combinatorial interpretation?
![Page 132: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/132.jpg)
The nth Catalan number is
Cn =1
n + 1
(2n
n
).
Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .
TheoremWe have C0 = 1 and, for n ≥ 1, Cn =
∑n−1i=0 CiCn−i−1.
Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.For example, let
Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.TheoremFor n ≥ 0 we have Cn = #Dn.
Define the fibocatalan numbers to be
Cn,F =1
Fn+1
(2n
n
)F
.
It is not hard to show Cn.F ∈ N for all n. Lou Shapiro asked: Canone find a combinatorial interpretation?
![Page 133: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/133.jpg)
The nth Catalan number is
Cn =1
n + 1
(2n
n
).
Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .
TheoremWe have C0 = 1 and, for n ≥ 1, Cn =
∑n−1i=0 CiCn−i−1.
Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.
For example, let
Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.TheoremFor n ≥ 0 we have Cn = #Dn.
Define the fibocatalan numbers to be
Cn,F =1
Fn+1
(2n
n
)F
.
It is not hard to show Cn.F ∈ N for all n. Lou Shapiro asked: Canone find a combinatorial interpretation?
![Page 134: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/134.jpg)
The nth Catalan number is
Cn =1
n + 1
(2n
n
).
Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .
TheoremWe have C0 = 1 and, for n ≥ 1, Cn =
∑n−1i=0 CiCn−i−1.
Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.For example, let
Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.
TheoremFor n ≥ 0 we have Cn = #Dn.
Define the fibocatalan numbers to be
Cn,F =1
Fn+1
(2n
n
)F
.
It is not hard to show Cn.F ∈ N for all n. Lou Shapiro asked: Canone find a combinatorial interpretation?
![Page 135: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/135.jpg)
The nth Catalan number is
Cn =1
n + 1
(2n
n
).
Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .
TheoremWe have C0 = 1 and, for n ≥ 1, Cn =
∑n−1i=0 CiCn−i−1.
Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.For example, let
Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.TheoremFor n ≥ 0 we have Cn = #Dn.
Define the fibocatalan numbers to be
Cn,F =1
Fn+1
(2n
n
)F
.
It is not hard to show Cn.F ∈ N for all n. Lou Shapiro asked: Canone find a combinatorial interpretation?
![Page 136: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/136.jpg)
The nth Catalan number is
Cn =1
n + 1
(2n
n
).
Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .
TheoremWe have C0 = 1 and, for n ≥ 1, Cn =
∑n−1i=0 CiCn−i−1.
Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.For example, let
Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.TheoremFor n ≥ 0 we have Cn = #Dn.
Define the fibocatalan numbers to be
Cn,F =1
Fn+1
(2n
n
)F
.
It is not hard to show Cn.F ∈ N for all n. Lou Shapiro asked: Canone find a combinatorial interpretation?
![Page 137: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/137.jpg)
The nth Catalan number is
Cn =1
n + 1
(2n
n
).
Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .
TheoremWe have C0 = 1 and, for n ≥ 1, Cn =
∑n−1i=0 CiCn−i−1.
Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.For example, let
Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.TheoremFor n ≥ 0 we have Cn = #Dn.
Define the fibocatalan numbers to be
Cn,F =1
Fn+1
(2n
n
)F
.
It is not hard to show Cn.F ∈ N for all n.
Lou Shapiro asked: Canone find a combinatorial interpretation?
![Page 138: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/138.jpg)
The nth Catalan number is
Cn =1
n + 1
(2n
n
).
Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .
TheoremWe have C0 = 1 and, for n ≥ 1, Cn =
∑n−1i=0 CiCn−i−1.
Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.For example, let
Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.TheoremFor n ≥ 0 we have Cn = #Dn.
Define the fibocatalan numbers to be
Cn,F =1
Fn+1
(2n
n
)F
.
It is not hard to show Cn.F ∈ N for all n. Lou Shapiro asked: Canone find a combinatorial interpretation?
![Page 139: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/139.jpg)
References.
1. Andrews, George E. The theory of partitions. Reprint of the1976 original. Cambridge Mathematical Library.CambridgeUniversity Press, Cambridge, (1998).
2. Benjamin, A. T., and Plott, S. S. A combinatorial approach toFibonomial coefficients, Fibonacci Quart. 46/47 (2008/09),7–9.
3. Gessel, I., and Viennot, G. Binomial determinants, paths, andhook length formulae. Adv. in Math. 58 (1985), 300–321.
4. Knuth, Donald E. Subspaces, subsets, and partitions. J.Combinatorial Theory Ser. A 10 (1971) 178–180.
5. MacMahon, Percy A. Combinatory Analysis. Volumes 1 and2. Reprint of the 1916 original. Dover, New York, (2004).
6. Sagan, Bruce E., and Savage, Carla D. Combinatorialinterpretations of binomial coefficient analogues related toLucas sequences. Integers 10 (2010), A52, 697–703.
7. Stanley, Richard P. Enumerative combinatorics. Volume 1.Second edition. Cambridge Studies in Advanced Mathematics,49. Cambridge University Press, Cambridge, (2012).
![Page 140: Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the](https://reader035.fdocuments.us/reader035/viewer/2022062917/5ed919c26714ca7f47692683/html5/thumbnails/140.jpg)
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