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!

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.
