Notation
$[x^k]f(x) =$ the coeffcient of $x^k$ in the power series expansion of $f(x)$
The Full Question
If I roll a die numbered from $1$ to $6$ twelve times, how many ways can they sum to $30$? (Use generating functions.
My Work
We can represent all the possible die rolls with the polynomial $f(x)=(x+x^2+x^3+x^4+x^5+x^6)^{12}= (x\frac{1-x^6}{1-x})^{12}=x^{12}(1-x^6)^{12}(\frac{1}{1-x})^{12}$
Let $a(x) = x^{12}(1-x^6)^{12},b(x) = (\frac{1}{1-x})^{12},c(x) = a(x)\times b(x)$.
Since $c(x)$ is our generating function and is also the product of two other functions, then the coefficient of $x^{30}$ in the power series expansion of $c(x)$ is our answer. We can use our theorem for the coefficient of the product of two power series to obtain this coefficient.
$[x^{30}]c(x) = \sum_{j=0}^{30}[x^j]a(x)[x^{30-j}]b(x)$
I know that $[x^{30-j}]b(x) = \binom{12+(30-j)-1}{30-j}$
My Question
Does anyone know how I can find the coeffcient of $a(x)$ I feel I'm very close to solving the problem, just this last part I'm having trouble with.
The function $a(x) = x^{12}(1 - x^5)^{12}$ is the product of $x^{12}$ and $(1 - x^5)^{12}$. Ok, so this isn't anything special so far. But when you have powers of differences, you can use the binomial theorem. In particular, $$ (a + b)^n = \sum_{k \geq 0} {n \choose k} a^{n - k} b^k,$$ so in particular, we have that $$ (1 - x^5)^{12} = \sum_{k \geq 0} {12 \choose k} (-x^5)^k.$$ This has a $0$th, $5$th, $10$th, ..., $60$th power of $x$. You can add $12$ to the powers of $x$ to reaccount for the thus omitted $x^{12}$ term.