Question

If ${{}^{n}C}_{12}={{}^{n}C}_8$, find ${{}^{n}C}_{17},{{}^{22}C}_{11}$

Answer 1

$\frac{n!}{(n-12)!(12!)}=\frac{n!}{(n-8)!(8!)}$

By solving above equation we get,

$(n-8)(n-9)(n-10)(n-11)=12\times 11\times 10\times 9$

So by comparing, we get $n-8=12\Rightarrow n=20$

Now,

${}^{20}{C}_{17}=\frac{20!}{3!\times 17!}=1140$

and,

${}^{22}{C}_{11}=705432$