Variation on Vandermonde's identity

How can you show that $$ \binom{2n}{n}^2 = \sum_{m=0}^{n} \binom{2n}{2m} \binom{2m}m \binom{2n-2m}{n-m} $$?

I was fooling around with random walks, and apparently both expressions are supposed to be the number of unweighted random walks on a 2D square lattice of $2n$ steps that ends at the origin.

Whole thing reminds me of Vandermonde's identity but I can't it. Any hints?

Solution 1:

$$\binom{2n}{n}^2 = \sum_{m=0}^{n} \binom{2n}{2m} \binom{2m}m \binom{2n-2m}{n-m}$$ Consider the right hand side terms. $$ \binom{2n}{2m} \binom{2m}m \binom{2n-2m}{n-m} = \frac{(2n)!}{(2m)!(2n-2m)!}\frac{(2m)!}{m!m!} \frac{(2n-2m)!}{(n-m)!(n-m)!} = \frac{(2n)!}{m!m!(n-m)!(n-m)!} = \frac{(2n)!}{n!n!}\frac{n!}{m!(n-m)!}\frac{n!}{m!(n-m)!}={2n\choose n}{n\choose m}{n \choose m} $$ Then the summation becomes $$ \sum_{m=0}^{n} \binom{2n}{2m} \binom{2m}m \binom{2n-2m}{n-m} = {2n\choose n}\sum_{m=0}^{n}{n\choose m}^2 $$ From here onwards, there is proof in the following page.

Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$

Solution 2:

On the right-hand side, $2m$ of the $2n$ steps are horizontal, $m$ of the $2m$ horizontal steps go right, and $n-m$ of the $2n-2m$ vertical steps go up.

For the left-hand side, turn the lattice by $\pi/4$ and independently choose $n$ out of $2n$ steps that go right and and $n$ out of $2n$ steps that go up.