<s>[INST] Generating functions for finding the coefficients I am new to the field of combinatorics and I recently came across a problem where it was asked to find the number of integer solutions to ${c_1 + c_2 + c_3 + c_4=20  }$ where ${c_i\ge 0}$ for all ${1\le i\le4}$ with ${c_2}$ and ${c_3}$ being even integers.
Using the generating function, we know that the solution would be the coefficient of ${x^{20}}$ in the expansion of ${(1 +x+x^2+x^3+...)^2*(1+x^2+x^4+x^6+...)^2}$ which is equivalent to: ${(1-x)^{-2}*(1-x^2)^{-2}}$
Now, we know that the first part could be solved by computing:${(-1)^{20}\dbinom{-2}{20} = \dbinom{2+20-1}{20}=21}$
Since we know that:  ${(1+x^m)^n = \dbinom{n}{0} + \dbinom{n}{1}x^m + \dbinom{n}{2}x^{2m}+...+\dbinom{n}{n}x^{nm}}$
Would it be correct to assume that the coefficient of ${x^{20}}$ in ${{(1-x^2)^{-2}}}$ would be:  ${\dbinom{-2}{10}}$
 [/INST] Yes, the coefficient of $x^{20}$ in $(1-x^2)^{-2}$ is the coefficient of $y^{10}$ in $(1-y)^{-2}$ which is $(-1)^{10}\binom{-2}{10}=\binom{-2}{10}$.
As shown in this answer,
$$
\binom{-2}{10}=(-1)^{10}\binom{11}{10}=11
$$

Computation of the Coefficients
$$
\begin{align}
(1-x)^{-2}(1-x^2)^{-2}
&=\sum_{j=0}^\infty(-1)^j\binom{-2}{j}x^j\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^{2k}\\
&=\sum_{j=0}^\infty\sum_{k=0}^{\lfloor j/2\rfloor}(-1)^{j-k}\binom{-2}{j-2k}\binom{-2}{k}x^j\\
&=\sum_{j=0}^\infty\sum_{k=0}^{\lfloor j/2\rfloor}\binom{j-2k+1}{j-2k}\binom{k+1}{k}x^j\\
&=\sum_{j=0}^\infty\sum_{k=0}^{\lfloor j/2\rfloor}(j-2k+1)(k+1)x^j\\
&=\sum_{j=0}^\infty\left[(j+1)\binom{\lfloor j/2\rfloor+2}{2}-4\binom{\lfloor j/2\rfloor+2}{3}\right]x^j\\
&=\sum_{j=0}^\infty\frac{j+3}{48}\left[\left(2j^2+12j+13\right)+(-1)^j3\right]x^j
\end{align}
$$
using the fact that $\lfloor j/2\rfloor=\frac{2j-1+(-1)^j}4$.
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