Find the coefficient of $x^{24}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^8$ I'm not sure how to go about doing this. Do I find the ways to add up to 24 using the exponents with repetition? Is the multinomial theorem useful here? I also have a feeling that generating functions might be useful here, but I can't see how. Any help would be appreciated.
 A: $$(1+x+\dots+x^5)^8=\left(\frac{1-x^6}{1-x}\right)^8=(1-x^6)^8(1-x)^{-8}$$
Using the binomial theoerem,
$$
(1-x^6)^8=\sum_{k\ge0}(-1)^k\binom{8}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-8}=\sum_{k\ge0}(-1)^k\binom{-8}{k}x^k=\sum_{k\ge0}\binom{8+k-1}{k}x^k
$$
Thus, when we convolve the above two generating functions, the $x^{24}$ coefficient is
$$
\binom{8}{0}\binom{8+24-1}{24}-\binom{8}{1}\binom{8+18-1}{18}+\binom{8}{2}\binom{8+12-1}{12}\\
-\binom{8}{3}\binom{8+6-1}{6}+\binom{8}{4}\binom{8+0-1}{0}
$$
Addendum: If $a(x)=\sum_{n\ge0}a_nx^n$ and $b(x)=\sum_{n\ge 0}b_nx^n$, then the $x^{24}$ coefficient of $c(x)=a(x)b(x)$ is 
$$
\sum_{k=0}^{24}a
_kb_{n-k}
$$
The final answer I wrote then comes from setting $a(x)=(1-x^6)^8$, $b(x)=(1-x)^{-8}$, and realizing that $a_k=0$ unless $k$ is a multiple of 6, so the above can be rewritten
$$
\sum_{\ell=0}^{4}a
_{6\ell}b_{n-6\ell}
$$
A: $\bf{My\; Solution::}$ Let $S = 1+x+x^2+x^3+........+x^5......(1)$
Multiply both side by $x\;,$ We get
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;xS = x+x^2+x^3+..............x^6........(2)$
Now Subtract $(1)$ and $(2)\;,$ we get
$\Rightarrow \displaystyle S(1-x) = 1-x^6\Rightarrow S = \frac{(1-x^6)}{(1-x)}$
So we have to find Coeff. of $x^{24}$ in $\displaystyle (1-x^6)^8\cdot (1-x)^8$
Now Using Binomial Theorem, For $(+ve)$ Integral Index 
$\displaystyle (1-y)^n = \binom{n}{0}-\binom{n}{1}y+\binom{n}{2}y^2+..............+(-1)^n\binom{n}{n}y^n$
and Using Binomial Theorem, For $(-ve)$  Integral Index
$\displaystyle (1-y)^{-n} = 1+ny+\frac{n(n+1)}{2}y^2+\frac{n(n+1)(n+2)}{3}y^3+.........$ 
