Question

If ${}^{2n+1}{\text{P}}_{n-1}$ : ${}^{2n-1}{\text{P}}_n$ = 3 : 5 then 'n' is

• A. 2
• B. 4
• C. 6
• D. 7

Answer 1

The correct option is B 4

Given :- $(2n+1)P_{n-1}:(2n-1)P_n=3:5$
$\Rightarrow 5.(2n+1)P_{n-1}=3.(2n-1)P_n$
$\Rightarrow 5.\frac{(2n+1)!}{(n+2)!}=3.\frac{(2n-1)!}{(n-1)!}$
$\Rightarrow 10(2n+1)=3(n+2)(n+1)$
$\Rightarrow 3n^2-11n-4=0$
$\Rightarrow n=4$