Combinatorial proof of $\sum_{i=0}^n {n \choose i}^2={2n \choose n}$ by using a binary string of length $2n$ [duplicate]

Question

I was wondering if there is any combinatorial proof of $$\sum_{i=0}^n {n \choose i}^2={2n \choose n}$$ It is obvious that the RHS of the equation counts the number of binary strings of length $2n$ with exactly $n$ 1's, but how about the LHS?

Answer 1

Suppose that we have $n$ girls and $n$ boys, and we want to pick $n$ people out of them. One way is $\large{2n \choose n}$. Another way is to pick $i$ girls, and $n-i$ boys, which is $${n \choose i}\times {n\choose n-i}={n \choose i}^2.$$ Summation over $i$ gives LHS.

Answer 2

Vandermonde's identity $$\sum_{i=0}^{k}\binom {m}{i}\binom {n}{k-i}=\binom {m+n}{k}$$ and $$\binom{n}{k-i}=\binom{n}{n-k+i}$$ Set $k=m=n$

Proof

Suppose a committee consists of $m$ men and $n$ women. In how many ways can a subcommittee of $k$ members be formed? The answer is $$\binom{m+n}{k}$$ The answer is also the sum over all possible values of $k$, of the number of subcommittees consisting of $k$ men and $k-i$ women: $$\sum_{i=0}^{k}\binom {m}{i}\binom {n}{k-i}$$