Generating function and combinatorics

Question

I am studying now the concept of generating function, and I have a solved question in my book which I don't understand, completely. There it is:

What is the number of options to roll 10 different dice, so the sum
of the results will be 25?

Now for the solution:

f(x) = (x+x^2+x^3+x^4+x^5+x^6)^10

I believe this is because we have 6 options in each dice, and 10 different dice.

=x^10(1+x+x^2+x^3+x^4+x^5)^10

Here the professor took out the Common factor x^10 (I think by taking out x^6 and some more)

=x^10(1-x^6)^10 * (1+x+x^2....)^10

The "...." sign is a sign for an infinite arithmetic progression or until x^5? I don't know.

Now we have to find the "base" of x^25, which is the basic of a generatic function or so I believe. Now he makes so permutations here:

C(10+15-1,15) - 10*C(10+9-1,9) + C(10,2)*C(10+3-1,3) =
=C(24,9)-10*C(18,9)+36*C(6,3)=
=1,307,504 - 486,200+720 = 822,040

I believe it has to be related to this concept, that the book shows for Derivative. For example, f'(x)=n/(1-x)^(n+1); f''(x)=n(n+1)/(1-x)^n+2 Now he shows for the k-th Derivative that: 1/k! * f'(x) in the k-th time, if we set x=0:

n(n+1)....(n+k-1) / k! = C(n+k-1,k)

I am sorry for the long, not so understandable question, but It's really hard for me to understand his methods.

Thank you!

Answer 1

Between your third gray and fourth gray, you should recognize that $1+x+x^2+x^3+x^4+x^5=\frac{1-x^6}{1-x}$, so you have $f(x)=x^{10}\left(\frac{1-x^6}{1-x}\right)^{10}$. Then $\frac 1{1-x}=1+x+x^2+x^3 \dots$ where the dots go on forever, getting you to the fourth gray. I can't help beyond there.

Answer 2

Since you have 10 dice, and each die can be any integer from 1 to 6, your generating function is

$f(x)=(x+x^2+x^3+x^4+x^5+x^6)^{10}=(x(1+x+x^2+x^3+x^4+x^5))^{10}=x^{10}(1+x+x^2+x^3+x^4+x^5)^{10}$ $=\displaystyle x^{10}\big(\frac{1-x^6}{1-x}\big)^{10}=x^{10}(1-x^6)^{10}(1-x)^{-10}.$

Now we have to find the coefficient of $x^{25}$, and we can use the Binomial Formula in the second factor and the formula $(1-x)^{-n}=\sum_{k=0}^{\infty}\binom{n-1+k}{k}x^k$ in the third factor.

This gives $f(x)=x^{10}(1-10x^6+45x^{12}+\cdots)\big(\sum_{k=0}^{\infty}\binom{9+k}{k}x^k\big)$, so the coefficient of $x^{25}$ will be the coefficient of $x^{15}$ in $(1-10x^6+45x^{12}+\cdots)\big(\sum_{k=0}^{\infty}\binom{9+k}{k}x^k\big)$, which is

$\;\;\binom{24}{15}-10\binom{18}{9}+45\binom{12}{3}=831, 204.$