Prove $\sum_ {k=0} ^n {n \choose k}^2 = {2n \choose n} $? Give a proof of the following identity using a double-counting argument: 
$\sum_ {k=0} ^r {m \choose k} {n \choose r - k} = {{m+n} \choose r} $ 
Then using this result, derive the following special case from it. 
$\sum_ {k=0} ^n {n \choose k}^2 = {2n \choose n} $ ?
For the first one, the method that I have is starting with the right hand side:


*

*Assume that k is a small number, which is less than r and m.

*We choose r from m + n can be divided into 2 parts:


*

*Choose k from m.

*Choose r - k from n.



My thinking is we can choose k from m and them choose r - k (because we only need to choose r) from n.
I'm stuck with the next part and not sure if my proof for the first part is right or not.
 A: Your overall idea for the first part is more or less right. Combinatorial arguments like this tend to flow more smoothly/logically if you think with a more concrete example. A popular idea seems to be committee selections.
If you want to reframe the first problem in such terms, think of it like this. Say we have $m$ men and $n$ women from which to pick a committee of $r$ members. Then on the one hand, we could pick $r$ from the entire group of $m+n$ people, thus getting $C(m+n,r)$ possible selections. 
Alternatively, we could pick $k$ men then and $r-k$ women (note that these sum to $r$). Well, what values could $k$ be? We could be $0$ men, $1$ man, $2$ men, and so on, up to $r$ men. Thus, combining all this, we would get the summation on the left-hand side since you have to sum over the valid values of $k$ to account for each case.

As for the second one, just take $n=r=m$. The result should appear immediately. Don't forget the symmetry property of the binomial coefficient:
$$\binom n k = \binom{n}{n-k}$$
