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Section 6 . 1: Generating Function Models
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Example
Suppose Caleb has a collection of 10 baseball cards. How manydifferent ways can Caleb select 6 different cards from his collection?
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Example
Suppose Caleb has a collection of 10 baseball cards. How manydifferent ways can Caleb select 6 different cards from his collection?This is asking how many ways are there to select a subset of 6cards out of a set of 10 cards. The answer is clearly,C (10 , 6) = 106 .In section 5. 5 you saw that 106 is the coefficient of x 6 in theexpansion of (1 + x )10
Such expressions are useful for generating solutions to questionsinvolving combinatorics, and so they have been called generatingfunctions. Our goal is to begin to understand what a generatingfunction is, how to write a generating function and how to use itto solve combinatorial problems.From http://math.illinoisstate.edu/day/courses/old/305/contentgeneratingfunctions.html
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Example - continued
How is (1 + x )10 a generating function for the problem involving Calebsbaseball cards?
Applications of generating functions take advantage of the additiveproperty of exponents: i.e. x 3x 4 = x 3+4 = x 7
Think of the ten factors in (1 + x )10 associated with each of Calebsten cards.Think of the expression (1 + x ) as representing:
(EXCLUDE THIS CARD + INCLUDE THIS CARD)
When we choose 1 from a factor as we expand we can think of thisas saying EXCLUDE THIS CARD. When we choose x from a factoras we expand we can think of this as saying INCLUDE THIS CARD.Thus in the expansion
(1 + x )10 = 1 +101
x +102
x 2 + +109
x 9 + x 10
the term 106 x 6 = 210 x 6 indicates that there are 210 different ways
to select 6 of the phrases INCLUDE THIS CARD.3 /14
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How many ways are there to select a four-lettercombination from the set {A , B , C } if A can be includedat most once, B can be included at most twice and C atmost three times?
One approach is to use cases:
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How many ways are there to select a four-lettercombination from the set {A , B , C } if A can be includedat most once, B can be included at most twice and C atmost three times?
One approach is to use cases:
Because the set must contain four letters, there must be at leastone C . If there is only one C , then there must be two B s. So theset is {C , B , B , A}What if there are two C s?
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How many ways are there to select a four-lettercombination from the set {A , B , C } if A can be includedat most once, B can be included at most twice and C atmost three times?
One approach is to use cases:
Because the set must contain four letters, there must be at leastone C . If there is only one C , then there must be two B s. So theset is {C , B , B , A}What if there are two C s?{C , C , B , B } or {C , C , B , A}What if there are three C s?
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How many ways are there to select a four-lettercombination from the set {A , B , C } if A can be includedat most once, B can be included at most twice and C atmost three times?
One approach is to use cases:
Because the set must contain four letters, there must be at leastone C . If there is only one C , then there must be two B s. So theset is {C , B , B , A}What if there are two C s?{C , C , B , B } or {C , C , B , A}What if there are three C s?{C , C , C , B } or {C , C , C , A}From http://math.illinoisstate.edu/day/courses/old/305/contentgeneratingfunctions.html
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How many ways are there to select a four-lettercombination from the set {A , B , C } if A can be includedat most once, B can be included at most twice and C atmost three times?
How can we use a generating function to model this?
We could use a polynomial to express each letters possibilities:(1 + A) represents A occurring 0 times or 1 time. Notice thatA0 = 1 .(1 + B + B 2) represents B occurring 0, 1 or 2 times.(1 + C + C 2 + C 3) represents C occurring 0, 1, 2, or 3 times. Then
the expansion of (1 + A)(1 + B + B 2)(1 + C + C 2 + C 3) will listfor us all the ways we can create k element sets within therestrictions of the problem.From http://math.illinoisstate.edu/day/courses/old/305/contentgeneratingfunctions.html
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How many ways are there to select a four-lettercombination from the set {A , B , C } if A can be includedat most once, B can be included at most twice and C atmost three times?
Then the expansion of (1 + A)(1+ B + B 2)(1+ C + C 2 + C 3) will list forus all the ways we can create k element sets within the restrictions of theproblem.
For example, one term in the expansion is AB 2C 2 and it represents a veelement set with one A, two B s and two C s.The problem doesnt ask us to list all the possibilities so the followinggenerating function is more efficient for nding the answer.
(1 + x )(1 + x + x 2)(1 + x + x 2 + x 3)
When we expand this we get
1 + 3 x + 5 x 2 + 6 x 3 + 5 x 4 + 3 x 5 + x 6
The coefficient on the x 4 term tells us the number of ways to create fourelement sets under the restrictions of this problem.From http://math.illinoisstate.edu/day/courses/old/305/contentgeneratingf unctions.html
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Some Denitions
Suppose ar is the number of ways to select r objects in a certainprocedure. Then g (x ) is a generating function for ar if g (x ) hasthe polynomial expansion
g (x ) = a0 + a1x + a2x 2 + + ar x r + + an x n
If the function has an innite number of terms, it is called a powerseries.
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Binomial Expansion
In sec 5. 5 you saw that
(1 + x )n = 1 +n1
x +n2
x 2 + +nr
x 2 + +nn
x n .
This means g (x ) = (1 + x )n
is the generating function fora r = C (n , r ) which is the number of ways to select a subset of r items from a set of n items.Determining the coefficient of x r in (1 + x )n is the same ascounting the number of different formal products with r x s and
(n r ) 1s. This is equivalent to counting the number of allsequences of r x s and (n r ) 1s. There are clearly C (n , r ) waysthis can occur.
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Formal Products
In the expansion of (1 + x + x 2)4, where we let 1 = x 0 the set of all formal products will be sequences of the form
x 0x x 2
x 0x x 2
x 0x x 2
x 0x x 2
All formal products can be written as x e 1 x e 2 x e 3 x e 4 with 0 e i 2.
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Example
If we want to nd the coefficient of x 5 in the expansion of (1 + x + x 2)4, this is equivalent to nding the number of differentformal products such that the sum of their exponents is ve.
e 1 + e 2 + e 3 + e 4 = 5 0 e i 2
This is the same as selecting ve objects from four types with atmost two objects of each type.
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Building a Generating Function
Suppose we have two apples, three nectarines and four plums. We
want to build a generating function for the number of ways toselect k pieces of fruit when we must select at least one of eachfruit. This could be modeled as the number of integer solutions to
e 1 + e 2 + e 3 = k 1 e i i + 1
Let (A + A2) represent we choose one apple or two apples.Let (N + N 2 + N 3) represent we choose one nectarine, twonectarines or three nectarines.Let (P + P 2 + P 3 + P 4) represent we choose one plum, two plums,three plums or four plums.Then (A + A2)(N + N 2 + N 3)(P + P 2 + P 3 + P 4) gives us agenerating function or a symbolic series representing all possiblefruit selections under the given constraints.
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Example
Take ( A + A2)(N + N 2 + N 3)(P + P 2 + P 3 + P 4) and replace A, N and P with x . This gives you
(x + x 2)(x + x 2+ x 3)(x + x 2+ x 3+ x 4) = x 3+3 x 4+5 x 5+6 x 6+5 x 7+3 x 8+ x 9
We can easily see that there are six ways to select six pieces of fruit under these restrictions.
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Example
Suppose we have two apples, three nectarines and four plums. We
built a generating function for the number of ways to select k pieces of fruit when we must select at least one of each fruit.Suppose an apple costs 40 cents, a nectarine costs 40 cents and aplum costs 20 cents. How should we change our generatingfunction so that the coefficients of x n in the resulting expression is
the number of fruit selections that cost n cents?
(x + x 2)(x + x 2 + x 3)(x + x 2 + x 3 + x 4)
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Example
Suppose we have two apples, three nectarines and four plums. We
built a generating function for the number of ways to select k pieces of fruit when we must select at least one of each fruit.Suppose an apple costs 40 cents, a nectarine costs 40 cents and aplum costs 20 cents. How should we change our generatingfunction so that the coefficients of x n in the resulting expression is
the number of fruit selections that cost n cents?
(x + x 2)(x + x 2 + x 3)(x + x 2 + x 3 + x 4)
(x 40
+ x 80
)(x 40
+ x 80
+ x 120
)(x 20
+ x 40
+ x 60
+ x 80
)= x 100+ x 120 +3 x 140+3 x 160+4 x 180+4 x 200+3 x 220+3 x 240+ x 260+ x 280
So we can see that there are four ways to select $2. 00 worth of fruit, taking at least one of each kind.
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References
Applied Combinatorics, 6th ed by Alan Tucker
Introductory Combinatorics, 3rd ed by Kenneth P. Bogart http://math.illinoisstate.edu/day/courses/old/305/contentgeneratingfunctions.html
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