Determining the number of solutions using Generating functions I have the equation
$u_1 + u_2 + ... + u_5 = 24$
with the restrictions
$1 \le u_i \le 7, i = 1,...5$
I've managed to work up to the point of finding the coefficient of $x^{24}$ in
$(X+X^2+...X^7)^5 = X^5(1+X+...X^6)^5$
I know my final answer will need to be in binomial form similar to this $\binom{10}{5}$
However I am unsure where to go, or if this is even correct. Any help would be greatly appreciated.
 A: Everything is fine with your generating function. It is convenient  to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.

We obtain
  \begin{align*}
[x^{24}]&(x+x^2+\cdots+x^7)^5\\
&=[x^{24}]x^5(1+x+\cdots+x^6)^5\\
&=[x^{19}]\left(\frac{1-x^{7}}{1-x}\right)^5\tag{1}\\
&=[x^{19}](1-x^7)^5\cdot\frac{1}{(1-x)^5}\\
&=[x^{19}](1-5x^{7}+10x^{14})\sum_{n=0}^\infty\binom{-5}{n}(-x)^n\tag{2}\\
&=\left([x^{19}]-5[x^{12}]+10[x^5]\right)\sum_{n=0}^\infty\binom{n+4}{4}x^n\tag{3}\\
&=\binom{23}{4}-5\binom{16}{4}+10\binom{9}{4}\tag{4}\\
&=1015
\end{align*}

Comment:


*

*In (1) we use the geometric series formula and apply the rule
\begin{align*}
[x^{p-q}]A(x)=[x^p]x^qA(x)
\end{align*}

*In (2) we use the binomial series expansion and expand $(1-x^7)^5$. We skip terms with exponent $21$ and greater since they do not contribute to $[x^{19}]$.

*In (3) we use the linearity of the coefficient of operator, apply again the rule as in (1) and we use the binomial identity $$\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$$

*In (4) we select the coefficients accordingly.
