How to find the coefficient without a calculator? I wanted to solve this question without using a calculator.
Question: The number of non-negative integer solutions to
$$3x+y+z=24$$
By creating generating functions you have to find the coefficient of $x^{24}$ in the expression: $$\left(\frac{1}{1-x}\right)^{2}\left(\frac{1}{1-x^3}\right)$$
Using the theory I know about now, I would just split the problem into smaller parts adding all the combinations together while using the extended binomial theorem. But this takes a lot of time and I was wondering if there is a faster/easier way to find coefficients in terms of multiple generating functions by hand? If so, what are some recommended places to read about it?

Pen, paper, and effort is what is required outside of a computer to calculate higher terms in expansions. In this case
\begin{align}
\frac{1}{(1-x)^2} &= \sum_{n=0}^{\infty} (n+1) \, x^n \\
\frac{1}{1-x^3} &= \sum_{n=0}^{\infty} x^{3 n}
\end{align}
for which the product of these series gives
$$ P = \frac{1}{(1-x)^2 (1-x^3)} = \sum_{r=0}^{\infty} (r+1) \, x^r \times \sum_{m=0}^{\infty} x^{3 m}. $$
Expanding to powers of $x^{24}$ gives
\begin{align}
P &= (1 + 2 x + 3 \, x^2 + \cdots + 25 \, x^{24})(1 + x^3 + \cdots + x^{24}) \\
&= \cdots + (1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 ) \ x^{24} + \cdots \\
&= \cdots + 108 \, x^{23} + 117 \, x^{24} + 126 \, x^{25} +  \cdots.
\end{align}
The resulting value is then $[x^{24}] P = 117$.
