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

[Albert1]

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

 

[kaliprasad]

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

Spoiler

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

 

[Albert1]

Spoiler

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

nice solution !