Albert1
- 1,221
- 0
prove:
$\sum_{k=0}^{n} \binom{n}{k}^2
=\dfrac{(2n)!}{(n!)^2}$
$\sum_{k=0}^{n} \binom{n}{k}^2
=\dfrac{(2n)!}{(n!)^2}$
Last edited:
Albert said:prove:
$\sum_{k=0}^{n} \binom{n}{k}^2
=\dfrac{(2n)!}{(n!)^2}$
nice solution !kaliprasad said: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