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Combinatorics CSC 172 SPRING 2002 LECTURE 11 Assignments With Replacement Example: Passwords Are...
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Transcript of Combinatorics CSC 172 SPRING 2002 LECTURE 11 Assignments With Replacement Example: Passwords Are...
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Combinatorics
CSC 172
SPRING 2002
LECTURE 11
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Assignments With Replacement
Example: PasswordsAre there more strings of length 5 built from three
symbols or strings of length 3 built from 5 symbols?
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In general
We are given n “items”, to each we must assign one of k “values”Each value may be used any number of timesLet W(n,k) be the number of ways
How many different ways may we assign values to the items?“Different ways” means that one or more of the items get different values.
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Inductive Definition
Basis:
W(1,k) == k
Induction:
If we have n+1 items, we assign the first one in one of k ways and the remaining n in W(n,k) ways
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Recurrence
W(1,k) = k
W(n+1,k) = k * W(n,k)
T(1) = k
T(n) = k * T(n-1)
Easy expansion
T(n) = kn
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Example: Are there more strings of length 5 built from three symbols or strings of length 3 built from 5 symbols?
Length 5, from 3 “values” = {0,1,2}
“items” are the 5 positions
N = 35 = 243
Length 3, from 5 “values” = {0,1,2,3,4}
“items” are the 3 positions
N = 53 = 125
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Permutations
Example: “SCRABBLE”
- start with 7 letters (tiles)
- how many different ways can you arrange them
- we don’t care about the word’s legality
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Scrabble
We can pick the first letter to be any of the 7 tilesFor each possible 1st letter, there are 6 choices of
second lettersOr, 7*6 = 42 possible two letter prefixesSimilarly, for each of the 42, there are 5 choices of
the third letter. 42 * 5 = 210, and so onTotal choices = 7*6*5*…*1 = 7! = 5040
In general, there are n! permutations of n items.
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Ordered Selections
Suppose we want to begin Scrabble with a 4 letter word? How many ways might we form the word from our 7 distinct tiles?
For each possible 1st letter, there are 6 choices of second letters 7*6 = 42 possible two letter prefixes
For each of the 42, there are 5 choices of the third letter. 42 * 5 = 210
For each of the 120, there are 4 choices of the third letter. 210 * 4 = 840
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In general
(n,m), the nmber of ways to pick a sequence of m things out of n
== n*(n-1)*(n-2)*…*(n-m+1)
== n!/(n-m)!
1*...*)1(*)(
1*...*)(*)1(*...*)2(*)1(*),(
mnmn
mnmnnnnmn
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Combinations
Suppose we give up trying to make a word and want to throw 4 of our 7 tiles back in the pile? How many different ways can we get rid of 4 tiles?
Ordered selection 4!/(7-4!) = 840
However, we don’t care about order.
So, how many ways are there to order 4 items?
4! = 24
840/24==35
== 7!/((7-4)!4!) = 35
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In general
“n choose m”
!)!(
!
mmn
n
m
n
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Recursive Definition for n choose mWe want to choose m things out of n, we can either
take or reject the first item.
If we take the first, then we can take the rest by choosing m-1 of the remaining n-1
We can do this in (n-1) choose (m-1) ways
OTOH, if we reject the first item, then we can get the rest by choosing m of the remaining n-1
We can do this in (n-1) choose m ways
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Inductive Definition
Basis: for all n there is only one way to choose all or
none of the elements
Induction:
for 0 < m < n
10
n
nn
m
n
m
n
m
n 1
1
1
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Proving Inductive Definition = Direct Definition
What is the induction parameter?
Zero for the basis case
decreases in the inductive step
Complete induction on m(n-m)
m
n
m
nmnc
1
1
1),(
Prove: c(n,m) = n!/((n-m)!m!)
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Basis
If m(n-m) == 0, then either m == 0 or m == n
If m == 0,
then n!/(n-m)!m! == n!/n! = 1 = c(n,0)
If m == n,
then n!/(n-m)!m! == n!/n! = 1 = c(n,n)
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Induction
By definition c(n,m) = c(n-1,m-1) + c(n-1,m)
Assume,c(n-1,m-1) = (n-1)!/((n-m)!(m-1)!)
c(n-1,m) = (n-1)!/((n-m-1)!m!)
Add the left sides == c(n,m), by definition
Add the right sides == n!/(n-m)!m!, clearly
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)!)!1((
)!1(
))!1()!((
)!1(
mmn
n
mmn
n
!)!1()!1()!(
)!1()!()!1(!)!1()!1(
mmnmmn
mmnnmmnn
!)!1()!(
)!()!1()!1()!1(
mmnmn
mnnmmnn
!)!(
)()!1()!1(
mmn
mnnmn
!)!(
)!1()!1()!1(
mmn
nmnnmn
!)!(
!
!)!(
)!1(
mmn
n
mmn
nn
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Orders With Some Equivalent ItemsIn real life, we play Scrabble with duplicate letters
Suppose you draw {S,T,A,A,E,E,E} at star, how many 7-letter “words” can you make.
Similar to permutations, but now, some are indistinguishable, because of duplicates
Trick: we can mark the letters to make them distinguishable S,T,A1,A2,E1,E2,E3
Then we get 7!=5040 ways
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How much “sameness”
But some order are the same
E3TA1E1SA2E2 == E3TA2E1SA1E2
The two As can be ordered in 2! = 2 ways
The three Es can be ordered in 3! = 6 ways
So, the number of different words is
7!/2!3! = 540/(2*6) = 420
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In general
The orders of n items with groups i1,i2,…,ik equivalent items is
n!/(i1!i2!..ik!)
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Items into Bins
Suppose we throw 7 dice (6 sided). How many outcomes are there?
Place each of 7 items into one of 6 bins The tokes are the dice
The bins are then number of dice
Putting the second token into bin 3 means that the second die shows 3
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TrickImagine 5 markers, denoted “*” that represent
separation between bins, and 7 tokens “T” that represent dice
The string “*TT**TTT*T*T” corresponds to no 1s, two 2s, no 3s, three 4s, a 5 and a 6
How many such strings?
“orders with identical items”
12 items, 5 of type “*”, 7 of type “T”
12!/5!7! = 792
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In general
The number of ways to assign n items to m bins
The number of orders of n-1 markers and n tokens
n
mn
nm
mn 1
!)!1(
)!1(
We are picking n out of the possible n+m-1positions for the tokens
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Several kinds of items into Bins
Order in bin can be either important (queues) or not
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Several kinds of items, order within bin unimportant
If we have items of k colors, with ij items of the jth color, then the number of distinguishable assignments into m bins is:
j
jk
j i
im 11
Example: 4 red dice, 3 blue dice, 2 green dice6 bins
176,1482
7
3
8
4
9
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Order important
If there are m bins into which ij items are placed and order within the bin matters
Imagine m-1 markers separating the bins and ij tokens of the jth type
By “orders with identical items”
jkj im
mn
1)!1(
)1(