Simplify the expression - sum of squares of binomial coefficients

Question

I need to simplify this thing: $$\sum_{k=0}^n{n\choose k}^2$$
I spent more than an hour thinking about it and the set of working solutions that sprang to my mind was unfortunately still empty. Then, I resorted to a book and came across this formula - the "Cauchy identity": $${m+n \choose k} = \sum_{s=0}^k{m \choose s}{n \choose k-s}$$ This already rang a bell - I return to the previous problem:
$$\sum_{k=0}^n{n \choose k}^2 = \sum_{k = 0}^n {n\choose k}{n \choose n-k}$$ Now, I assumed that $m=n$ (There was no restriction in the identity) and that $s=n$ again, no restriction.
Therefore, I got this simplification, being simultaneously my final answer:
$${2n \choose n}$$
Do you think that this solution works? If so, I am not trying to conceal the fact that I find this solution very unnatural and I would have never solved it were it not for the book. Is there a simpler way to do this? Maybe something on the combinatorial level?

Answer 1

A combinatorical proof could be like this:

Assume you have $2n$ balls and you want to choose $n$ of them. One way is straight-forward choosing $n$ from $2n$ by $\binom{2n}{n}$ ways

Also you can group them in 2 n-groups and choose k from the first one and n-k from the second one which gives us $\sum \binom{n}{k}\binom{n}{n-k}$ but since $\binom{n}{k}=\binom{n}{n-k}$ you could also write it as following: $$\sum \binom{n}{k}\binom{n}{n-k} = \sum \binom{n}{k}\binom{n}{k} = \sum \binom{n}{k}^2$$

So overall we have

$$ \binom{2n}{n} = \sum \binom{n}{k}^2 $$

Answer 2

Consider $(1+x)^{n} \cdot (1+x)^{n} = (1+x)^{2n}$. Now what is the coefficient before $x^{n}$?