Vandermonde identity corollary $\sum_{v=0}^{n}\frac{(2n)!}{(v!)^2(n-v)!^2}={2n \choose n}^2$

Question

I am trying to prove this identity: $$\sum_{v=0}^{n}\frac{(2n)!}{(v!)^2(n-v)!^2}={2n \choose n}^2$$ I think this identity (corollary of Vandermonde identity): $${n\choose 0}^2+{n\choose 1}^2+{n\choose 2}^2+\cdots+{n\choose n}^2={2n \choose n}$$ is applicable for solving it. Please give me some hints. thank you

Answer 1

Vandermonde's identity says

$${(m+n)\choose r} = \sum_{k=0}^{r} {m\choose k}{n\choose (r-k)}$$

Put $m = n$, $n = n$, and $r = n$.

You get$$ {(2n)\choose n} = {n\choose 0}{n\choose n}+{n\choose 1}{n\choose (n-1)}+\cdots + {n\choose n}{n\choose n}$$

$${(2n)\choose n} = {(n)\choose 0}^2+{(n)\choose 1}^2+\cdots+{(n)\choose n}^2\tag1$$

From your expression when expanded gives

$$\frac{(2n)!}{n!^2} + \frac{(2n)!}{((n-1)!)^2} +\frac{(2n)!}{(2!(n-2)!)^2} +\cdots+\frac{(2n)!}{(n!n!)^2}$$

Muptiply and divide by $(n!)^2$

$$\frac{(2n)!}{n!^2}\left(\frac{n!^2}{n!^2}+\frac{n!^2}{1!.(n-1)!)^2}+\frac{n!^2}{(2!.(n-2)!)^2}+\cdots+ \frac{n!^2}{n!^2}\right)$$

This reduces to $${2n\choose n}\left({n\choose 0}^2+{n\choose 1}^2+\cdots+{n\choose n}^2\right)$$

From (1) we get $${2n\choose n}.{2n\choose n}$$

$$({2n\choose n})^2$$