Prove that $\sum_{r=0}^n({n\choose r})^2={2n\choose n }$

Question

Prove that

$\sum_{r=0}^n{n\choose r}^2={2n\choose n }$

I don't know how to prove such probems. Any proof by combinatorics?

Answer 1

On the right hand side is the number of ways to choose n people from 2n people. If we split the 2n people into two groups size n, A and B, say-

We can choose 0 people from group A, n people from group B, 1 from group A, n-1 people from group B, etc.

Adding up all possibilities, this gives rise to: $\sum^{n}_{k=0}\binom{n}{k}\binom{n}{n-k}=\binom{2n}{n}$

But, $\binom{n}{k}=\binom{n}{n-k}$, so therefore, $\binom{2n}{n}=\sum^{n}_{k=0}\binom{n}{k}^2$

Another proof can be found by considering $(1+x)^{2n}=(1+x)^n(1+x)^n$, and comparing coefficients of $x^n$ on both sides.