Question

Determine $n,\mathrm{if}{}^{2\mathrm{n}}\mathrm{C}_2:{}^{\mathrm{n}}\mathrm{C}_2::9:2$.

• A. $5$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is A

$5$

Explanation for correct option:

Given, ${}^{2\mathrm{n}}\mathrm{C}_2:{}^{\mathrm{n}}\mathrm{C}_2::9:2$

$$\begin{gathered} \Rightarrow \frac{\left({\displaystyle \frac{\left(2n\right)!}{\left(2n-2\right)!2!}}\right)}{\left({\displaystyle \frac{n!}{\left(n-2\right)!2!}}\right)}=\frac{9}{2}\left(\because {}^{n}C_r=\frac{n!}{\left(n-r\right)!r!}\right) \\ \Rightarrow \frac{\left({\displaystyle \frac{2n\left(2n-1\right)\left(2n-2\right)!}{\left(2n-2\right)!}}\right)}{\left({\displaystyle \frac{n\left(n-1\right)\left(n-2\right)!}{\left(n-2\right)!}}\right)}=\frac{9}{2} \\ \Rightarrow \frac{2\left(2n-1\right)}{\left(n-1\right)}=\frac{9}{2} \\ \Rightarrow 8n-4=9n-9 \\ \Rightarrow n=5 \end{gathered}$$

Hence, correct answer is option $\left(A\right)$.