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!
 A: 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.
A: Computing the generating function:
$$
\begin{align}
&\sum_k\sum_l\sum_i\sum_j\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j}x^ly^k\\
&=\sum_k\color{#C00}{\sum_l}\sum_i\sum_j\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\color{#C00}{\binom{4}{k-l+i+j}x^{l+4-k-i-j}}x^{k+i+j-4}y^k\\
&=\color{#C00}{(1+x)^4}\sum_k\sum_i\sum_j\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}x^{k+i+j-4}y^k\\
&=(1+x)^4\color{#090}{\sum_k}\sum_i\sum_j\binom{2}{i}\binom{2}{j}\color{#090}{\binom{2}{k-i-j}x^{k-i-j}y^{k-i-j}}x^{2i+2j-4}y^{i+j}\\
&=(1+x)^4\color{#090}{(1+xy)^2}\color{#00F}{\sum_i\sum_j\binom{2}{i}\binom{2}{j}x^{2i+2j-4}y^{i+j}}\\
&=\frac{(1+x)^4(1+xy)^2\color{#00F}{(1+x^2y)^4}}{\color{#00F}{x^4}}\\
&=\color{#CCC}{\frac1{x^4}+\frac{4+2y}{x^3}+\frac{6+12y+y^2}{x^2}+\frac{4+28y+12y^2}x}\\
&+\left(1+32y+44y^2+4y^3\right)+x\left(18y+76y^2+28y^3\right)+x^2\left(4y+69y^2+76y^3+6y^4\right)\\
&+x^3\left(32y^2+104y^3+32y^4\right)+x^4\left(6y^2+76y^3+69y^4+4y^5\right)\\
&+x^5\left(28y^3+76y^4+18y^5\right)+x^6\left(4y^3+44y^4+32y^5+y^6\right)\\
&\color{#CCC}{+x^7\left(12y^4+28y^5+4y^6\right)+x^8\left(y^4+12y^5+6y^6\right)+x^9\left(2y^5+4y^6\right)+x^{10}y^6}
\end{align}
$$
where the terms not in the requested range have been grayed out.
A: I would add some comments following the given solution. First at all we need a variable change $l'=l+4$ to bring the GF into the real world.

Suppose we have to fill the structure above with $l'$ identical white balls and $k$ identical black balls, white in the upper row, black in the lower row. Then there is a rule that says every structured bin either is empty or is full. 
Thus we get $\binom2i$ for filling the first section, $\binom2j$ for the second, $\binom2{k-i-j}$ for the green section. For the fourth section, we have $l'-2i -2j - (k-i-j) = l'-i-j-k = l + 4 - i-j-k$ hence the binomial $\binom4 {i+j+k-l} $ 
here are my comments :


*

*such structures that could be named ''partial surjective function'' missed the twelve-fold Rota way train or other expansions and they are less studied.  

*the blue summamnds could be grouped in only one section with only one parameter, but the section is split.

*the l' parameter is shifted from reality exactly with 4 as to be well hidden in the binomial expression. 

*then we have a range for l', to place 4..10 balls in 14 slots.
Gives these I would say, someone have done his mile to produce this tough structure and problem.
