Binomial Coefficient $\sum_{k=0}^n {2n \choose 2k}{2k \choose k}{2n-2k \choose n-k} $

Question

I am trying to find the sum
$$\sum_{k=0}^n {2n \choose 2k}{2k \choose k}{2n-2k \choose n-k} $$
I know the closed-form answer and I could derive it based on the form of the answer in terms of the coefficient of some expansion but I am curious to know how to solve this problem without knowing the answer.

Answer 1

Your summation can be written:

$$S=\sum_{k=0}^n \dfrac{(2n)!}{(2n-2k)!(2k)!}\dfrac{(2k)!}{(k)!(k)!}\dfrac{(2n-2k)!}{(n-k)!(n-k)!}$$

By cancellation and introduction of $(n)!(n)!$:

$$S=\sum_{k=0}^n \dfrac{(2n)!}{(n)!(n)!}\left(\dfrac{(n)!}{(k)!(n-k)!}\right)^2$$

$$S=\binom{2n}{n}\sum_{k=0}^n \binom{n}{k}^2$$

Finally, using using a classical result:

$$S=\binom{2n}{n}^2$$

Edit: I just found the very same question here with, in particular, an interesting combinatorial answer.