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.

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.
