# Proof of a combinatorics identity [duplicate]

I am currently studying probability and am having difficulty understanding this question:
Prove that $$\binom{2n}{n} = \sum_{k=0}^n\binom{n}{k}^2$$ from Jim Pitman's Probability (Appendix I, question vi). So far I have tried writing the right hand side in different ways: expressing it in its factorial form and have tried to implement the identity $$\binom{n}{k} = \binom{n-1}{k-1}+\binom{n-1}{k}$$ but have not gained any new ground. If any one has an algebraic proof, or even a simple combinatorics proof that is intuitive and used with an example that would be preferable. Thanks!

## marked as duplicate by Namaste, ccorn, Arnaud Mortier, J.-E. Pin, Ethan BolkerJun 13 '18 at 17:29

$$(1+x)^{2n}=(1+x)^n(1+x)^n$$ $$\sum_{k=0}^{2n}{2n\choose k}x^k=\left(\sum_{k=0}^n{n\choose k}x^k\right)^2=\sum_{k=0}^{2n}x^k\sum_{i=0}^k{n\choose i}{n\choose k-i}$$
The coefficient of $x^n$ is
$${2n\choose n}=\sum_{i=0}^n{n\choose i}{n\choose n-i}=\sum_{i=0}^n{n\choose i}^2$$
To choose $n$ people out of $2n$ people, first split the $2n$ people arbitrarily into $2$ groups. Now, take $k$ people from the first group and $n-k$ people from the second group. If $k$ is fixed, one can do this in ${n\choose k} {n\choose n-k}={n\choose k}^2$. Summing over $k$ to account for all possible $k$ gives the result.