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?
 A: 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 $[x^{24}] P = 117$.
A: Here's an alternative approach that uses stars and bars instead of generating functions.  Condition on the number of solutions with $x=k$, which reduces to the number of nonnegative integers solutions to $y+z=24-3k$:
\begin{align}
\sum_{k=0}^8 \binom{24-3k+2-1}{2-1}
&= \sum_{k=0}^8 (25-3k) \\
&= 9\cdot 25 - 3\sum_{k=0}^8 k \\
&= 9\cdot 25 - 3\cdot 9 \cdot \frac{0+8}{2} \\
&= 9(25 - 12) \\
&= 117
\end{align}
