Simplify summation with factorial and binomial coefficients I would like to know how to simplify the following summation:
$$\sum_{p=0}^n\quad n!\frac{(2p)!}{(p!)^2}\frac{(2(n-p))!}{((n-p)!)^2}$$
Which rules should I use to simplify it?
Thanks!
 A: The sum is just:
$$
n! \sum_{0 \le p \le n} \binom{2 p}{p} \binom{2 (n - p)}{n - p}
$$
Vandermonde's convolution says:
$$
\sum_{0 \le k \le r} \binom{m}{k} \binom{n}{r - k} = \binom{m + n}{r}
$$
so your mistery sum is just:
$$
n! \binom{2 p + 2 (n - p)}{n} = n! \binom{2 n}{n!} = (2 n)^{\underline{n}}
$$
(here $x^{\underline{n}} = x (x - 1) \ldots (x - n + 1)$).
A: You may start with the generating function of $\displaystyle \frac{(2k)!}{(k!)^2}$ which is :
$$f(x):=\sum_{k=0}^\infty\ \frac{(2k)!}{(k!)^2}x^k=\sum_{k=0}^\infty\ \binom{2k}{k}x^k$$
but this is the well known central binomial generating function :
$$f(x)=\frac 1{\sqrt{1-4x}}$$
and compute the square of $f(x)$ so that the coefficients of the 'diagonal terms' $\;\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{(n-k)}x^kx^{n-k}\,$ will be your sums for increasing values of $n$
(ignoring the 'constant' factor $n!$) :
$$f(x)^2=\frac 1{1-4x}=\sum_{n=0}^\infty\ 4^n\;x^n$$
From this we deduce simply :
$$\boxed{\displaystyle n!\sum_{p=0}^n\ \frac{(2p)!}{(p!)^2}\frac{(2(n-p))!}{((n-p)!)^2}=\ n!\,4^n}$$
