Question

$Findthevalueofnif$
$\frac{\left(2n\right)!}{3!\left(2n-3\right)!}:\frac{n!}{2!(n-2)!}=44:3.$

Answer 1

$\frac{\left(2n\right)!}{3!(2n-3)!}:\frac{n!}{2!(n-2)!}=\frac{44}{3}$

$\frac{\left(2n\right)!}{3!(2n-3)!}\times \frac{2!(n-2)!}{n!}=\frac{44}{3}$

$\frac{\left(2n\right)(2n-1)(2n-2)(2n-3)!}{3.2\times (2n-3)!}\times \frac{2!\times (n-2)!}{n(n-1)(n-2)!}=\frac{44}{3}$

$\frac{(2n-1)(2n-2)}{n-1}=22$

$2(2n-1)=22$

$2n-1=11$

$2n=12$

$\Rightarrow n=6$.