Question

If $2{\times }^n{C}_5=9\times {}^{n-2}C_5$, then the value of $n$ will be?

• A. $7$
• B. $10$
• C. $9$
• D. $5$

Answer 1

The correct option is B

$10$

Explanation for the correct option:

Find the value of $n$:

Given, $2{\times }^n{C}_5=9\times {}^{n-2}C_5$

$\Rightarrow$ $2\times \frac{n!}{5!\times \left(n-5\right)!}=9\times \frac{(n-2)!}{(n-7)!5!}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$\Rightarrow$ $\frac{2(n-1)\left(n\right)}{(n-6)(n-5)}=9$

$\Rightarrow$ $2n^2-2n=9(n^2-11n+30)$

$\Rightarrow$$7n^2-97n+270=0$

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

$\because n\in N$

$\mathbf{\therefore }\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{10}$

Hence, Option ‘B’ is Correct.