Solve the following

$2 \times{ }^{n} C_{5}=9 \times{ }^{n-2} C_{5}$

i. e $2 \times \frac{n !}{5 !(n-5) !}=9 \times \frac{(n-2) !}{5 !(n-2-5) !}$

i. $\mathrm{e} \frac{2 \times n !}{5 !(n-5) !}=\frac{9 \times(n-2) !}{5 !(n-7) !}$

i. e $\frac{2 \times n(n-1)(n-2) !}{5 !(n-5)(n-6)(n-7) !}=\frac{9 \times(n-2) !}{5 !(n-7) !}$

i. e $2 \times[n(n-1)]=9(n-5)(n-6)$

i. e $2 n^{2}-2 n=9\left[n^{2}-11 n+30\right]$

$2 n^{2}-2 n=9 n^{2}-99 n+270$

i. e $7 n^{2}-97 n+270$

$D=9409-4(7)(270)$

$=9409-7560=1849$

Using, formula $\frac{-b \pm \sqrt{D}}{2 a}=\frac{97 \pm \sqrt{1849}}{2 \times 7}=\frac{97 \pm 43}{14}$

$=10, \frac{27}{7}$

Since $n \in N, \Rightarrow n=10$