Algebraic Way to Prove $\sum_{r=0}^{n} \frac{(2n)!}{((n-r)!)^2(r!)^2} = \binom{2n}{n}^2$

Question

$\sum_{r=0}^{n} \frac{(2n)!}{((n-r)!)^2(r!)^2} = \binom{2n}{n}^2$

This identity can be easily proven by mathematical induction.

However, is there any algebraic way to sum it up to $\binom{2n}{n}^2$?

Answer 1

Observe that $$\frac{(2n)!}{((n-r)!)^2(r!)^2}=\frac{(n!)^2}{((n-r)!)^2(r!)^2}\frac{(2n)!}{(n!)^2}=\binom{2n}{n}\binom{n}{r}^2$$ Therefore, $$\sum_{r=0}^{n} \frac{(2n)!}{((n-r)!)^2(r!)^2}=\sum_{r=0}^{n}\binom{2n}{n}\binom{n}{r}^2=\binom{2n}{n}\sum_{r=0}^{n}\binom{n}{r}^2=\binom{2n}{n}^2$$

Where $$\sum_{r=0}^{n}\binom{n}{r}^2=\binom{2n}{n}$$ follows from Vandermonde's identity.