Find the number of ways to distribute 10 pieces of candy using this generating function The question asks: 
a) Find a generating function for the number of ways to distribute identical pieces of candy to 3 children so that no child gets more than 4 pieces. Write this generating function in closed form, as a quotient of polynomials. b) Find the number of ways to distribute 10 pieces of candy using this generating function.
I figured out for part a that $(1 + x + x^2 + x^3 + x^4)^3$ is represented as the generating function:
$$f(x) = \left( \frac{1-x^5}{1-x} \right) ^3$$
But I don't know how to find the $a_{10}$ term. Please help. 
 A: 
Denoting with $[x^n]$ the coefficient of $x^n$ of a series we obtain
  \begin{align*}
\color{blue}{a_{10}}&\color{blue}{=[x^{10}]\left( \frac{1-x^5}{1-x} \right)^3}\\
&=[x^{10}]\frac{1-3x^5+3x^{10}-x^{15}}{(1-x)^3}\tag{1}\\
&=[x^{10}]\left(1-3x^5+3x^{10}\right)\sum_{j=0}^\infty \binom{-3}{j}(-x)^j\tag{2}\\
&=\left([x^{10}]-3[x^5]+3[x^0]\right)\sum_{j=0}^\infty \binom{j+2}{2}x^j\tag{3}\\
&=\binom{12}{2}-3\binom{7}{2}+3\binom{2}{2}\tag{4}\\
&=66-3\cdot 21+3\\
&\,\,\color{blue}{=6}
\end{align*}

Comment:


*

*In (1) we expand the numerator.

*In (2) we skip the terms of the numerator which do not contribute to $[x^{10}]$ and apply the binomial series expansion.

*In (3) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.

*In (4) we select the coefficients accordingly.
A: You need to manipulate first term of $(1-x^5)^3 (1-x)^{-3}$ in such way that would allow you to find the coefficients. The coefficients up to $x^{10}$, i.e. $a_{10}$, will reveal you the different ways you may distribute. 
Hint: Binomial theorem.
