Non-inductive, not combinatorial proof of $\sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ I've seen the identity $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ used here recently.

I checked for proofs here http://www.proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients

I couldn't have figured out the combinatorial proof by myself, and the inductive proof assumes you already know the answer...

So my question is : do you know how to prove directly through computation that $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ ?
Consider the identity $(1+x)^{2n}=(1+x)^n\cdot(1+x)^n$.  By the binomial theorem we have $\displaystyle(1+x)^n=\sum_{k=0}^n{n\choose k} x^k$, so multiplying out we compute the right hand side as $\displaystyle(1+x)^n\cdot(1+x)^n = \sum_{k=0}^{2n}\left(\sum_{j=0}^k{n\choose j}{n\choose k-j}x^k\right)$.  But the LHS is just $\displaystyle(1+x)^{2n} = \sum_{k=0}^{2n}{2n\choose k}x^k$; equating coefficients of $x^n$ we get $\displaystyle{2n\choose n}=\sum_{j=0}^n {n\choose j}{n\choose n-j}$.  Finally, using the identity ${n\choose j}={n\choose n-j}$ gives the desired result.
Note that we can divide the $2n$ objects in $2$ groups including $n$ elements each.

Then to choose $n$ elements out of $2n$ we can choose $0$ objects in the first group and $n$ in the second, or $1$ in the first and $n-1$ in the second and so on, i.e $$\binom{2n}{n} = \sum_{i = 0}^{2n}\binom{n}{i}\binom{n}{n-i} =\sum_{i = 0}^{2n}{\binom{n}{i}}^2 $$