Number of ways to choose $m$ boys and $k$ girls from $n$ boys and $n$ girls?

Question

Suppose there are $n$ boys and $n$ girls and we want to choose $m$ boys and $k$ girls such that $k \le m$. Then there are $\binom{n}{m} \binom{n}{k}$ ways to do it. Now, using counting in two ways, I want to prove that $$ \binom{n}{m} \binom{n}{k} = \sum_{j=0}^{k} \binom{n}{j} \binom{n-j}{m-j} \binom{n-m}{k-j}.$$

I am not able to figure out $j$ here, any hint to solve this problem? Because there are equal number of boys and girls, it is becoming difficult to understand that we are choosing boys or girls in the summation on RHS.

Answer 1

Hint: Imagine if you will that the boys and girls happen to be in brother-sister pairs.

Let $j$ represent the number of pairs where both siblings are taken for our resulting selection.

$~$

Pick which $j$ pairs are taken in $\binom{n}{j}$ ways. Then, pick the remaining necessary amount of boys noting that they may not be from those already selected pairs in $\binom{n-j}{m-j}$ ways. Finally, pick the remaining necessary amount of girls, noting they may not be siblings of any already selected boys (to avoid overcounting) in $\binom{n-m}{k-j}$ ways.