Solve the following


If $\frac{(2 n) !}{3 !(2 n-3) !}$ and $\frac{n !}{2 !(n-2) !}$ are in the ratio $44: 3$, find $n$.


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

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

$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)[(2 n-3) !]}{3(2 !)(2 n-3) !} \times \frac{2 !(n-2) !}{n(n-1)[(n-2) !]}=\frac{44}{3}$

$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{3} \times \frac{1}{n(n-1)}=\frac{44}{3}$

$\Rightarrow \frac{(2 n)(2 n-1)(2)(n-1)}{3} \times \frac{1}{n(n-1)}=\frac{44}{3}$

$\Rightarrow \frac{4(2 n-1) n(n-1)}{3} \times \frac{1}{n(n-1)}=\frac{44}{3}$

$\Rightarrow \frac{4(2 n-1)}{3}=\frac{44}{3}$

$\Rightarrow(2 n-1)=11$

$\Rightarrow 2 n=12$

$\Rightarrow n=6$

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