How is $\binom{2n}{n} \sum_{k=0}^{n}\binom{n}{k} \binom{n}{n-k}=(\binom{2n}{n})^{2}$

Question

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.

Answer 1

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,

$$\binom{n}{0}\times \binom {n}{n} + \binom{n}{1}\times \binom {n}{n-1} + \binom{n}{2}\times \binom {n}{n-2} + ... +\binom{n}{n}\times \binom {n}{0}$$

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