<s>[INST] What is the coefficient of $x^5$ in $(1+x+x^2+x^3+x^4+x^5)^{17}$? I figured that $(1+x+x^2+x^3+x^4)^{17} = (1-x^6)^{17}*(1-x)^{-17}$ but don't know what else to do. 
I would really appreciate any help
 [/INST] It is equivalent to the problem: how many ways can $5$ candies be distributed among $17$ children? Mathematically, it is:
$$x_1+x_2+\cdots +x_{17}=5, 0\le x_i\le 5, i=1,2,...,17.$$
The partitions of $5$ are:
$$\begin{align}\{5\} &\Rightarrow P(17,1)=\frac{17!}{16!}=17\\
\{4,1\} &\Rightarrow P(17,2)=\frac{17!}{15!}=272\\
\{3,2\}&\Rightarrow P(17,2)=\frac{17!}{15!}=272\\
\{3,1,1\}&\Rightarrow \frac{P(17,3)}{2!}=\frac{17!}{2\cdot 14!}=2040\\
\{2,2,1\}&\Rightarrow \frac{P(17,3)}{2!}=\frac{17!}{2\cdot 14!}=2040\\
\{2,1,1,1\}&\Rightarrow \frac{P(17,4)}{3!}=\frac{17!}{6\cdot 13!}=9520\\
\{1,1,1,1,1\}&\Rightarrow \frac{P(17,5)}{5!}=\frac{17!}{120\cdot 12!}=6188\\
\end{align}$$
Hence, the sum is: $20,349$.
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