Simplify $ \sum\limits_{i=0}^l{n\choose i}{m\choose l-i}$

Question

Is there a nice way to simplify $ \sum\limits_{i=0}^l{n\choose i}{m\choose l-i}$? I tried to tinker a bit with telescope sums but it did not get me nowhere...

Answer 1

The sum should be equal to $\binom{n+m}{l}$

One way to see this is to give a combinatorial proof of this. Let us consider the following scenario, we have a group of $n$ boys and $m$ girls, and we wish to choose $l$ people from the group of $n+m$ people. Thus, on the one hand, that value is $\binom{n+m}{l}$. On the other hand, we can choose $l$ people from the group by considering how many boys and girls we choose, if we choose no boys, then we need to pick $l$ girls, and there are $\binom{n}{0}\binom{m}{l}$ ways to do this, then if we choose $1$ boy we must have $l-1$ girls, and there will be $\binom{n}{1}\binom{m}{l-1}$, and continuing on if we choose $i$ boys, we must pick $l-i$ girls, and there will be $\binom{n}{i}\binom{m}{l-i}$. Thus, adding up all of these ways we have the way of choosing $l$ people from a group of $n+m$ people will be $\sum_{i=0}^l\binom{n}{i}\binom{m}{l-i}$