Question

If ${}^{2n+1}{P}_{n-1}:{}^{2n-1}{P}_n=3:5$ , then the value of $n$ is equal to :

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is A $4$

Given , ${}^{2n+1}{P}_{n-1}:{}^{2n-1}{P}_n=3:5$
$\Rightarrow \frac{(2n+1)!}{(n+2)!}\times \frac{(n-1)!}{(2n-1)!}=\frac{3}{5}$
$\Rightarrow \frac{(2n+1)\left(2n\right)}{(n+2)(n+1)n}=\frac{3}{5}$
$\Rightarrow 10(2n+1)=3(n^2+3n+2)$

$\Rightarrow 20n+10=3n^2+9n+6$
$\Rightarrow 3n^2-11n-4=0$
$\Rightarrow (3n+1)(n-4)=0$
$\Rightarrow n=4$ ..... [Since, $n$ can't be negative]