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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!

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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 : $$\boxed{\displaystyle n!\sum_{p=0}^n\ \frac{(2p)!}{(p!)^2}\frac{(2(n-p))!}{((n-p)!)^2}=\ n!\,4^n}$$

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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 $x^{\underline{n}} = x (x - 1) \ldots (x - n + 1)$).

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