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?
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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.
Emax
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Thanks, But how can I show that LHS actually counts the number of binary string of length 2n with exactly n 1's? – scott green Jan 22 '17 at 18:13
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@scottgreen You can assign "1" to the ones you pick and "0" to the others. – Emax Jan 22 '17 at 18:16
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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: $$\sum_{i=0}^{k}\binom {m}{i}\binom {n}{k-i}$$
Behrouz Maleki
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