Simplification of a Combinatoric problem (n k)^2

Question

How do I simplify the sum $$\sum_{k=0}^{n} {n \choose k} ^2?$$ I have tried, but the best I did was $$\sum_{k=0}^{n} {\frac{n!n!}{k!k!(n-k)!(n-k)!}}.$$

Answer 1

Here is a trick that reduces it. First we can say that $$\sum_{k=0}^n{n\choose k}^2 = \sum_{k=0}^n{n\choose k}{n\choose n-k}$$

Now let's think about what this means. ${n\choose k}{n\choose n-k}$ is the number of ways to pick $k$ things from a group of $n$, and then pick $n-k$ things from another group of $n$. So if we have a group of $2n$ things that we split it two, it is the number of ways to pick $k$ things from the first group and $n-k$ things from the second group. The crucial observation is that summing this over all k gives the total number of ways to pick $n$ things from the whole group! So the answer is $$\sum_{k=0}^n{n\choose k}^2 = {2n\choose n}$$