6.1 Generating Function Models (a+x)(a+x)(a+x) = aaa + aax + axa + axx xaa + xax + xxa + xxx

= 1a3 + 3a2x + 3ax2 + 1x3

(1+x)(1+x)(1+x) = 111 + 11x + 1x1 + 1xx

x11 + x1x + xx1 + xxx

= 1 + 3x + 3x2 + 1x3

= C(3,0) + C(3,1)x + C(3,2)x2 + C(3,3)x3

(1+x)3 is generating function for ar, the ways to select r objects from a set of 3 distinct objects.

If ar is the ways to do some combinatorial activity with r objects, then f(x) is the generating function for ar when f(x) = a0 + a1x + a2x2

. . . . + arxr +

E.g., (1+x)n = C(n,0) + C(n,1)x + C(n,2)x2 + . . C(n,n)xn is the generating function for ways to select r items from an n-set.

Or

Consider formal products yielding x5. The number of such formal products is the same as the number of integer solutions for the ei’s of

Example 1: Find the generating function for ar, the ways to select r balls from 3 green, 3 blue, and 3 gold balls.

As an integer-solution-of-equation problem, we have

e1+e2+e3= r, 0 ≤ ei ≤ 3 formal products xe1• xe2•xe3 = xr

Generating function (1 + x + x2 + x3 )3

Example 2: Find a gen. fn. for the number of ways to select r donuts from 5 chocolate, 5 strawberry, 3 lemon and 3 cherry donuts. As an integer-solution-of-equation problem, we have e1 + e2 + e3 + e4 = r, 0 ≤ e1, e2 ≤ 5, 0 ≤ e3, e4 ≤ 3

xe1•xe2•xe3•xe4 = xr

Gen. function (1 + x + x2 + x3+ x4 + x5)2(1 + x + x2 + x3 )2

Repeat with requiring at least one of each type. Now 1 ≤ ei – so (x + x2 + x3+ x4 + x5)2(x + x2 + x3 )2

Example 3: Use a gen. fn. to model the problem of counting all selections of 6 objects chosen from 3 types with repetition of up to 4 objects of each type. We build a standard gen. fn. and ask for its coefficient of x6. (1 + x + x2 + x3+ x4)3

Also build gen. fn. model with unlimited repetition. (1 + x + x2 + x3+ . . . . . .)3 Works for any finite value of r.

Example 4: Find a gen. fn. for ar, the ways to distribute r identical objects into 5 boxes with an even number not exceeding 10 in the first two boxes and between three and five in the other three boxes. We have to ‘customize’ our integer-solution-of-equations model for these particular available values for each box

e1 + e2 + e3 + e4 + e5= r, 0 ≤ e1, e2 even ≤ 10 and 3 ≤ e3,e4,e5≤ 5

(1 + x2 + x4 + x6+ x8 + x10)2(x3 + x4 + x5 )3

Binomial Identities

Set x = 1 in (1+x)n = Σ C(n,k)xn You get (2)n = Σ C(n,k)

Set x = 1 in (1-x)n = Σ C(n,k)(-x)n You get 0 = Σ (-1)nC(n,k) In other words, the sum of the odd terms cancels, i.e., equals, the sum of the even terms. So sum of odd terms = sum of even terms = ½(2n) = 2n-1

Differentiate both sides of (1+x)n = Σ C(n,k)xk yielding n(1+x)n-1 = Σ kC(n,k)x(k-1) and now set x=1 yielding n(2)n-1 = ΣkC(n,k)

Section 6.4 Exponential Generating Functions for arrangements with repetitionExample: Consider arrangements of length 4 using a’s, b’s, c’s with at least two a’sPossible patterns 4,0,0-- 4!/4!0!0! 3,1,0-- 4!/3!1!0! 3,0,1-- 4!/3!0!1! 2,2,0-- 4!/2!2!0! 2,1,1-- 4!/2!1!1! 2,0,2-- 4!/2!0!2!

We have one pattern for each solution to e1 + e2 + e3 = 4, e1 ≥ 2 (1)

So we do not want a count of one for each solution of (1), but rather each solution should have a count of 4!/e1!e2!e3!

Ordinary generating functions would count solutions to problems like (1) when order does not matter with(x2 + x3 + x4 . . .)(1 + x + x2 + x3 + )2

The exponential generating function for this problem is

The coefficient of xr will be the sum of all products

where e1 + e2 + e3 = 4, e1 ≥ 2.

If we divide the xr by r! and compensate by multiplying its coefficient by r!, the xr/r! term in g(x) becomes

e1 ≥ 2

Example 1: Find the exponential generating function for ar , the number of r arrangements without repetition of n objects. Ans: (1 + x/1!)n = (1 + x)n = S C(n,r)xr = S {n!/[(n-r)!r!]}xr

S {n!/(n-r)!}xr/r! or S P(n,r)(xr/r!) the ways to arrange r out of n objects.

Example 2: Find the exponential generating function for ar , the number of different arrangements of r objects chosen from 4 different types of objects with each type appearing at least two and at most five times. Ans: (x2/2! + x3/3! + x4/4! + x5/5!)4

Example3: Find the exponential generating function for ar , the number of ways to place r (distinct) people into three different rooms with at least one person in each room. (x/1! + x2/2! + x3/3! + . . . . . .)3 Repeat with an even number of people in each room. (x2/2! + x4/4! + x6/6! + . . . . . .)3