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.

$\bf{My\; Solution::}$ 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
$\displaystyle (1-y)^{-n} = 1+ny+\frac{n(n+1)}{2}y^2+\frac{n(n+1)(n+2)}{3}y^3+.........$ 
