1. Generating Functions
A robust tool to solve a wide variety of problems that deal with sequences, series, permutations, combinations, and more.
A generating function gives us a way to mathematically represent or encode such problems.
Definition
Purpose
Examples

2. Generating function A polynomial with either
Interesting choices for its coefficients
(useful for recurrence relations)
Interesting choices for its exponents
(useful for permutations/combinations)
Customary to write in ascending powers of x
Coefficients can correspond to terms of a sequence
Sequences have no mathematical operators. Writing as a polynomial provides a way to manipulate sequences.
Convenient if number of terms is infinite
We almost never substitute a value for x
 We are not interested in numerical value of polynomial: just manipulate its coef & exponents.

3. Purpose Our textbook shows how to solve simple:
Permutation & combinations
Recurrence relations
Using techniques that are narrowly defined.
Generating functions allow you to solve a much wider variety of problems in a systematic, less ad-hoc way.
Disadvantage: not very easy to use
Can be tedious
Requires techniques such as binomial expansion or partial fractions along the way
Can use Mathematica / MATLAB / Maple to help with some tedium, if desired.

4. Preliminary remarks How many terms do you get if you multiply…?
(a + b)(c + d)
(a + b)(c + d)(e + f)
(a + b + c)(d + e + f + g)
Notice that each term of the product tells you which number you selected inside each ( ).
If you replace each letter with “1”, you get # of terms!
Analogous concept: how many factors does a number have, e.g. 1200?

5. Combinations Consider the multiplication of (1+ax)(1+bx)(1+cx)
We need to create terms, where we select something from each factor:
Select the 1 or ax from (1+ax)
Select the 1 or bx from (1+bx)
Select the 1 or cx from (1+cx)
And you sum all 8 possibilities to get:
1 + (a+b+c)x + (ab + bc + ac)x2 + (abc)x3
What are the coefficients saying? 
If you take away the letters a,b,c, you get the answer to a different question: (1+x)3 = 1 + 3x + 3x2 + x3

6. continued A simple generating function for combinations is
(1+x)(1+x)(1+x)…(1+x) = (1+x)n
The coefficient of xr is the number of ways to choose r objects from a set of n distinct objects.
You are basically deciding which x’s to multiply to produce the xr.
What if the collection from which we are selecting has duplicates? Notice the contrast:
Select 2 objects from { A, B, C, D }
Select 2 objects from ( A, A, B, B ).

7. Duplicates
To select from ( A, A, B, B ) is like selecting from a set { A, B} with the ability to choose up to two of each!
Try this generating function:
(1 + ax + a2x2) (1 + bx + b2x2)
= 1 + (a+b)x + (a2 + ab + b2)x2 + (ab2 + a2b)x3 + a2b2x4.
If we just want the number of ways, we don’t need to specify “a” and “b”:
(1+x+x2) (1+x+x2) = 1 + 2x + 3x2 + 2x3 + x4.
What do the coefficients 1, 2, 3, 2, 1 tell us?
Try another problem like this.

8. Duplicates (2)
I have 10 A’s and 3 B’s. How many ways are there to select r objects, given that the number of A’s I can select must be a multiple of 5?
(1 + x5 + x10)(1 + x + x2 + x3)
= 1 + x + x2 + x3 + x5 + x6 + x7 + x8 + x10 + x11 + x12 + x13
What answers do you see in this polynomial?
Restriction on A’s: this problem can be stated as a coin problem with pennies and nickels! Exponent on x = total number of cents
We can easily generalize to different numbers plus more categories such as C’s, D’s, etc.
In this case it turned out all coefficients were 1. Not true in general.

9. Formulating
We need a separate factor for each distinct value we are selecting, or for each distinct category, die roll, digit in a number, etc.
Decide on what the exponent of x means. Possibilities:
Number of objects we select
The sum of the numbers selected
How many odd / even / multiple of 3 …
When done, coefficient of some term gives us our answer.
You want to know just how many, or a full enumeration?
More than one variable can be used if you need to keep track of more than one quantity about the selection.

10. Applications How to select r objects from a set
Allowing for repetition of the same element
Restrictions
Answer = coefficient of xr in some ( ) ( ) generating function.
Similar problems that are only superficially different
Coin problems
Dice problems

11. Review
What combination question would this generating function help solve?
(1 + ax + a2x2) (1 + bx) (cx + c2x2)
We can elegantly encode restrictions: maximum, minimum, even number, etc.
What if we just wanted the # of possibilities?
What if the a’s were unlimited?
The number of ways to choose coins to obtain c cents from pennies and nickels.
Think of the set { P, N } but the number of N’s you choose must be a multiple of 5.

12. Dice application How many ways can we roll a sum of 7 with 2 dice?
Formulation:
(x + x2 + x3 + x4 + x5 + x6) (x + x2 + x3 + x4 + x5 + x6)
= (x + x2 + x3 + x4 + x5 + x6)2
And we want the coefficient of x7.
We can match x with x6; x2 with x5; etc.
Thought process is parallel to how you might do with without generating functions.
The answer is 6.
Note that this question is equivalent to asking for coefficient of x5 in (1 + x + x2 + x3 + x4 + x5)2.

13. 3 dice
But, what if we had more than 2 dice? Generating functions “scale” very well.
Sum of 7 for 3 dice? No problem:
We want coefficient of x7 in (x + x2 + x3 + x4 + x5 + x6)3
Equivalent to asking for coefficient of x4 in
(1 + x + x2 + x3 + x4 + x5)3.
Inside ( ) we have a geometric series = (1 – x6)/(1 – x)
So (1 + x + x2 + x3 + x4 + x5)3 = (1 – x6)3(1 – x)–3
Let’s work out (1 – x6)3 and (1 – x)–3….

14. continued (1 – x6)3 = 1 – 3x6 + 3x12 – x18
We need to use the fact that the coefficients of (a+b)n are 1, n, n(n – 1)/2!, n(n – 1)(n – 2)/3!, n(n – 1)(n – 2)(n – 3)/4!, etc.
(1 – x6)3 = 1 – 3x6 + 3x12 – x18
(1 – x)–3 = 1 + 3x + (3)(4)/2! x2 + (3)(4)(5)/3! x3 + …
If we multiply these polynomials, how do we find a term containing x4?
x18, x12 and x6 can’t help us, because 2nd polynomial has only positive powers of x.
The only possibility is to multiply “1” from first polynomial by the x4 term from the 2nd polynomial.
Its coefficient is (3)(4)(5)(6)/4! = (5)(6) / 2 = 
In (1-x)^(-3), it turns out each term has even # of -. Can try a different sum or different # of dice.

15. Alternate calculation
Determining the coefficient of a single term can be phrased as a combination question.
The coefficient of x4 in (1 + x + x2 + x3 + x4 + x5)3 means we want 3 exponents to sum to 4, where each exponent ranges from 0 to 5.
In other words, e1 + e2 + e3 = 4 where each ei = 0,1,…,5.
How can 3 numbers sum to 4?
0+0+4, 0+1+3, 0+2+2, 1+1+2
In each case, decide which ei represents each number.
Total number of ways = = 15.

16. Ball in urn
Many combination questions are of the ball-in-urn type. We have n identical balls to place in r urns. The balls are identical, so we simply need to know how many wind up in each urn.
Basic approach: the generating function is of the form
(1 + x + x2 + x3 + …)r
and we want the coefficient of xn.
Highest term in each factor can simply be xn.
Distribute 12 bottles of wine among 4 tables…
What if each table must get 1+ bottle?
What if no table may get more than 4?
The first table must have at most 1 bottle, the 2nd must receive at least 1, the third must receive 0, 2 or 4; and the last must receive 2 or 3?

17. Practice formulating
A combination problem can be encoded as a generating function. Try these…
How many ways are there to select 4 objects from [ 5 A’s, 5 B’s, 5 C’s ]?
Now, what if we wanted an equal number of A’s and C’s?
How many ways can you make $1 change using pennies, nickels, dimes and/or quarters?
How many integers from 0 to 999 have a sum of digits of 12?
Formulating is more important than just computing an answer.

18. Two variables
Yes, it’s possible for a generating function to have 2 variables.
Why? Maybe you want to keep track of two quantities at the same time.
Example: (1, 1, 2, 3, 3, 4, 4, 5, 5, 6). How many ways to…
select r different numbers
select numbers that have a sum of r
select r numbers that have a sum of s.
Example: (0, 1, 2, 2, 3, 4)
(1 + x)(1 + xy)(1 + xy2 + x2y4)(1 + xy3)(1 + xy4)
Some of the terms include 3x3y5, 2x3y4, 3x2y4, x2y7

19. Review
The coefficient of x4 in (1 + x + x2)5 can answer what combinatorical question?
What if we had several dice, and…
You don’t have to roll all of them.
The dice do not all have the same number of sides.
You want to achieve a certain sum with a certain number of dice.
You want a sum of 31 with 4 6-sided, 2 8-sided, 2 12-sided and you must roll at least 1 of each.
Can you think of a question that would need 3 variables?
Using the Coefficient command in Wolfram Alpha
Example: Coefficient[(1+x+x^2)^10 (1+y)^4,x^5 y]