TITLE: Proving a binomial coefficient identity
QUESTION [5 upvotes]: I'm having some trouble with the following proof: $$\sum^k_{a=0} {{n}\choose{a}}{{m}\choose{k-a}} = {{n+m}\choose{k}}$$
I'm trying to prove this to learn a couple of things about the Pascal's triangle. I don't know where to start on the proof. I have tried expanding both sides using the binomial coefficient property but have had no luck.

REPLY [4 votes]: The RHS counts the number of ways of selecting a committee of $k$ people from $n$ women and $m$ men.  
The expression 
$$\binom{n}{a}\binom{m}{k - a}$$
counts the number of ways of selecting a committee of $k$ people consisting of $a$ women and $k - a$ men selected from selected from a group consisting of $n$ women and $m$ men.  Thus, the summation on the LHS counts all the ways of selecting a committee of $k$ people can be selected from $n$ women and $m$ men.