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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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