2021/100) prove: $\sum_{k=0}^{n} \binom{n}{k}^2 =\dfrac{(2n)!}{(n!)^2}$

Out of 2n objects n objects can be chosen in

$\dfrac{(2n)!}{(n!)^2}$ ways

Now let us make 2n objects into 2 groups of n objects each.

For picking n objects from the set we need k objects from 1st set and n-k from 2nd set and n varies from 0 to n so number of ways

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

The 2 above are same as it shows the number of ways in 2 different ways

So

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

Now as $\binom{n}{k} = \binom{n}{n-k}$ so we get the result