<s>[INST] a tough sum of binomial coefficients 
Find the sum: $$\sum_{i=0}^{2}\sum_{j=0}^{2}\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j},\space\space 0\leq k,l\leq 6$$

I know to find $\sum_{i=0}^{2}\binom{2}{i}\binom{2}{2-i}$, I need to find the coefficient of $x^2$ of $(1+x)^4$ (which is $\binom{4}{2}$). But I failed to use that trick here. Any help appreciated!
 [/INST] Here is a variant which could be seen as generalisation of OP's example. We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance
\begin{align*}
  [z^k](1+z)^n=\binom{n}{k}
  \end{align*}
and we also use Iverson brackets which are defined as
\begin{align*}
[[P(z)]]=\begin{cases}
1&\qquad P(z) \ \text{  true}\\
0&\qquad P(z) \ \text{ false}
\end{cases}
\end{align*}

We obtain for $0\leq k,l\leq 6$:
  \begin{align*}
\color{blue}{\sum_{i=0}^2}&\color{blue}{\sum_{j=0}^2\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j}}\\
&=\sum_{i=0}^2\binom{2}{i}\sum_{j=0}^2\binom{2}{j}[z^{k-i-j}](1+z)^2[u^{k-l+i+j}](1+u)^4\tag{1}\\
&=[z^k][u^{k-l}](1+z)^2(1+u)^4\sum_{i=0}^2\binom{2}{i}\left(\frac{z}{u}\right)^i\sum_{j=0}^2\binom{2}{j}\left(\frac{z}{u}\right)^j\tag{2}\\
&=[z^k][u^{k-l}](1+z)^2(1+u)^4\left(1+\frac{z}{u}\right)^4\tag{3}\\
&=[u^{k-l}]\left([z^k]+2[z^{k-1}]+[z^{k-2}]\right)\left(1+\frac{z}{u}\right)^4(1+u)^4\tag{4}\\
&=[u^{k-l}]\left(\binom{4}{k}u^{-k}+2\binom{4}{k-1}[[k\geq 1]]u^{1-k}\right.\\
&\qquad\qquad\quad\left.+\binom{4}{k-2}[[k\geq 2]]u^{2-k}\right)(1+u)^4\tag{5}\\
&=\left(\binom{4}{k}[u^{2k-l}]+2\binom{4}{k-1}[[k\geq 1]][u^{2k-l-1}]\right.\\
&\qquad\qquad\quad\left.+\binom{4}{k-2}[[k\geq 2]][u^{2k-l-2}]\right)(1+u)^4\\
&\,\,\color{blue}{=\binom{4}{k}\binom{4}{2k-l}[[2k\geq l]]+2\binom{4}{k-1}\binom{4}{2k-l-1}[[k\geq 1]][[2k\geq l+1]]}\\
&\qquad\qquad\quad\color{blue}{+\binom{4}{k-2}\binom{4}{2k-l-2}[[k\geq 2]][[2k\geq l+2]]}
\end{align*}

Comment:


*

*In (1) we apply the coefficient of operator twice.

*In (2) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(x)=[z^p]z^qA(z)$.

*In (3) we apply the binomial theorem twice.

*In (4) we expand $(1+z)^2$ and select the coefficient of $[z^k]$.
\begin{align*}
[z^k]&(1+z)^2\left(1+\frac{z}{u}\right)^4\\
&=[z^k](1+2z+z^2)\left(1+\frac{z}{u}\right)^4\\
&=\left([z^k]+2[z^{k-1}]+[z^{k-2}]\right)\left(1+\frac{z}{u}\right)^4\\
&=\left([z^k]+2[z^{k-1}]+[z^{k-2}]\right)\sum_{j=0}^4\binom{4}{j}\left(\frac{z}{u}\right)^j\\
&=[z^k]\sum_{j=0}^4\binom{4}{j}\left(\frac{z}{u}\right)^j
+2[z^{k-1}]\sum_{j=0}^4\binom{4}{j}\left(\frac{z}{u}\right)^j
+[z^{k-2}]\sum_{j=0}^4\binom{4}{j}\left(\frac{z}{u}\right)^j\\
\end{align*}

*In (5) we select the coefficients of $[z^{k-a}]$ in $\left(1+\frac{z}{u}\right)^4$ with $0\leq a\leq 2$. We use Iverson brackets to set terms to zero if the lower part of binomial coefficients is less than zero. We do a similar job with $[u^{k-l}]$ in the following lines.
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