Question

If ${}^{(2n+1)}{P}_{n-1}{:}^{(2n-1)}{P}_n=3:5$ then $n=$

• A. $4$
• B. $5$
• C. $6$
• D. $3$

Answer 1

The correct option is A $4$

Given equation is ${}^{2n+1}{P}_{n-1}{:}^{2n-1}{P}_n=3:5$

So,$\frac{\frac{(2n+1)!}{(2n+1-n+1)!}}{\frac{(2n-1)!}{(2n-1-n)!}}=\frac{3}{5}$

$\Rightarrow \frac{(2n+1)\left(2n\right)(2n-1)!}{(n+2)(n+1)n(n-1)!}\times \frac{(n-1)!}{(2n-1)!}=\frac{3}{5}$

Solving we get

$\Rightarrow \frac{2(2n+1)}{(n+1)(n+2)}=\frac{3}{5}$

$\Rightarrow 20n+10=3n^2+9n+6$

$\Rightarrow 3n^2-11n-4=0$

$\Rightarrow 3n^2-12n+n-4=0$

$\Rightarrow 3n(n-4)+1(n-4)=0$

$\therefore n=4$