Solve the following

Solution:

Given: $2 n_{C_{3}}: n_{C_{2}}=44: 3$

$\frac{2 n_{C_{3}}}{n_{C_{2}}}=\frac{44}{3}$

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

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

$\Rightarrow(2 n-1)(2 n-2)=22(n-1)$

$\Rightarrow 4 n^{2}-6 n+2=22 n-22$

$\Rightarrow 4 n^{2}-28 n+24=0$

$\Rightarrow n^{2}-7 n+6=0$

$\Rightarrow n^{2}-6 n-n+6=0$

$\Rightarrow n(n-6)-1(n-6)=0$

$\Rightarrow(n-1)(n-6)=0$

$\Rightarrow n=1$ or, $n=6$

Now, $n=1 \Rightarrow 2_{C_{3}}: 2_{C_{2}}=44: 3$

But, this is not possible.

$\therefore n=6$