<s>[INST] How to quickly find the $x^{24}$ term in this expansion? Is there a swift way to find the $x^{24}$ coefficient in the expansion of 
$$ \left(1-x^6\right)^{-2} \left(1-x^3\right)^{-1} \left(1-x\right)^{-1} $$
The general term of each bracket is $(r+1)x^{6r}$, $x^{3r}$ and $x^{r}$ respectively. 
 [/INST] The given rational function can be written as
$$(1+x^6+x^{12}+\ldots)(1+x^6+x^{12}+\ldots)(1+x^3+x^6+\ldots)(1+x^2+x^3+\ldots)\ .$$
The coefficient $c_{24}$ we are after therefore is the number of nonnegative integer solutions to
$$6k_1+6k_2+3k_3+k_4=24\ ,$$
or $6k_1+6k_2+3k_3\leq24$, which is the same as
$$2(k_1+k_2)+k_3\leq8\ .$$
If $k_1+k_2=:j\geq0$ each given value of $j$ ca be realized by $k_1$ and $k_2$ in $j+1$ ways, and then $k_3$ may take $8-2j+1$ different values. It follows that
$$c_{24}=\sum_{j=0}^4(j+1)\cdot(9-2j)=1\cdot9+2\cdot7+3\cdot5+4\cdot 3+5\cdot1=55\ .$$
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