I have been trying to make out how: $\binom{2n}{n} \sum_{k=0}^{n}\binom{n}{k} \binom{n}{n-k}=\binom{2n}{n}^{2}$ so in essence, showing that
$\sum_{k=0}^{n}\binom{n}{k} \binom{n}{n-k}=\binom{2n}{n}$
but I do not see this is the case, I mean in each summand, for a given $k$ I get,
$\frac{n!^{2}}{k!^{2}(n-k)!^{2}}$ I do not see how I can take out a common $\frac{2n!}{n!^{2}}$ out of each summand.
Any help is greatly appreciated.
Use the following two identities:
$$\binom{n}{k} = \binom{n}{n-k}$$
$$\sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}$$
For the proof of the second identity, consider the binomial expansion of $(1+x)^{2n}$ as
$$(1+x)^{2n} = (1+x)^{n} \times (1+x)^{n}$$
And now compare the coefficient of $x^n$ on both sides.
On the LHS it is simply $\binom{2n}{n}$.
On the RHS it will be,
$$\binom{n}{0}\times \binom {n}{n} + \binom{n}{1}\times \binom {n}{n-1} + \binom{n}{2}\times \binom {n}{n-2} + ... +\binom{n}{n}\times \binom {n}{0}$$
$$=\sum_{k=0}^{n} \binom{n}{k}^2$$
