Question

If ${}^{2n+1}{P}_{n-1}{:}^{2n-1}{P}_n=3:5$ find $n$.

Answer 1

Writing the given ratio in the fraction as:

$\frac{{}^{2n+1}{P}_{n-1}}{{}^{2n-1}{P}_n}=\frac{3}{5}$

i.e. $5\times {}^{2n+1}{P}_{n-1}=3\times {}^{2n-1}{P}_n$

$⟹\frac{5.(2n+1)!}{[2n+1-(n-1\left)\right]!}=\frac{3.(2n-1)!}{(2n-1-n)!}$ ... $[\because {}^n{P}_r=\frac{n!}{(n-r)!}]$
$⟹\frac{5(2n+1)2n(2n-1)!}{(n+2)!}=\frac{3(2n-1)!}{(n-1)!}$
$⟹\frac{10n(2n+1)}{(n+2)(n+1)n(n-1)!}=\frac{3}{(n-1)!}$
$\therefore 10(2n+1)=3(n+2)(n+1)$
$3n^2-11n-4=0$ $\Rightarrow (3n+1)(n-4)=0$
$\therefore n=-\frac{1}{3}$ or $n=4$