Question

Arrange the following values of n in ascending order.
$A{:}^n{P}_5{=}^n{P}_6\Rightarrow n=$
$B{:}^n{P}_{12}{=}^n{P}_8\Rightarrow n=$
$C{:}^n{C}_{(n-3)}=10\Rightarrow n=$
$D{:}^{(n+1)}{P}_5{:}^n{P}_6=1:2\Rightarrow n=$

• A. CABD
• B. CADB
• C. ACBD
• D. DBAC

Answer 1

The correct option is B CADB

A : ${}^n{P}_5{=}^n{P}_6$

$\Rightarrow \frac{n!}{(n-5)!}=\frac{n!}{(n-6)!}$

$\Rightarrow n-5=1$

$\Rightarrow n=6$

B : ${}^n{P}_{12}{=}^n{P}_8$

$\Rightarrow n=20$

$C{:}^n{C}_{n-3}{=}^n{C}_3=10$

$\Rightarrow n=5$

$D:\frac{(n+1)!}{(n-4)!}\times \frac{(n-6)!}{n!}=\frac{1}{2}$
$\Rightarrow \frac{n+1}{(n-4)(n-5)}=\frac{1}{2}$

$\Rightarrow n=9$

Hence, the ascending order of $n$ is $=$ CADB