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. $1$
• E. $5$

Answer 1

The correct option is A

$4$

Explanation of the correct answer:

The permutation of $r$ things from $n$ things can be given as ${}^{n}P_r$ such that

${}^n{P}_r=\frac{n!}{(n-r)!}$.

where $n$ and $r$ are positive integers.

It is given that

${}^{2n+1}P_{n-1}:{}^{2n-1}{P}_n=3:5$

It can be written as

$\frac{{}^{2n+1}{P}_{n-1}}{{}^{2n-1}C_n}=\frac{3}{5}$ …..(i)

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

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

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

By expanding the factorials, it can be written as

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

$\Rightarrow$$\frac{(2n+1)}{(n+2)(n+1)}=\frac{3}{10}$

$\Rightarrow$$10(2n+1)=3(n+2)(n+1)$ …..(ii)

Equation (ii) can be simplified as

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

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

$\Rightarrow$$n=4,\frac{-1}{3}$

Also $n$is always positive. Therefore,

$\mathit{n}\mathbf{=}\mathbf{4}$

Hence, Option (A) is correct.