Question

Find $r$ if $\left(i\right){}^5{P}_r=2{}^6{P}_{r-1}\left(ii\right){}^5{P}_r={}^6{P}_{r-1}$

Answer 1

$\left(i\right)$ Given$:{}^5{P}_r=2{}^6{P}_{r-1}$

${}^n{P}_r=\frac{n!}{(n-r)!}$
$\therefore \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-(r-1\left)\right)!}$
$\Rightarrow \frac{5!}{(5-r)!}=2\times \frac{6\times 5!}{(7-r)!}$
$\Rightarrow \frac{(7-r)(6-r)(5-r)!}{(5-r)!}=\frac{2\times 6\times 5!}{5!}$
$\Rightarrow (7-r)(6-r)=12$
$\Rightarrow 42-7r-6r+r^2=12$
$\Rightarrow r^2-13r+30=0$
$\Rightarrow r^2-3r-10r+30=0$
$\Rightarrow r(r-3)-10(r-3)=0$
$\Rightarrow (r-3)(r-10)=0$
$\therefore r=3$ or $r=10$
$\because r\le n\Rightarrow r\le 5$
$\therefore r=10$ is not possible.
$\Rightarrow r=3$.

$\left(ii\right)$ Given$:{}^5{P}_r{=}^6{P}_{r-1}$
$\Rightarrow \frac{5!}{(5-r)!}=\frac{6!}{(6-(r-1\left)\right)!}$
$\Rightarrow \frac{5!}{(5-r)!}=\frac{6\times 5!}{(7-r)!}$
$\Rightarrow \frac{5!(7-r)!}{(5-r)!}=6\times 5!$
$\Rightarrow \frac{(7-r)(6-r)(5-r)!}{(5-r)!}=\frac{6\times 5!}{5!}$
$\Rightarrow 42-7r-6r+r^2=6$
$\Rightarrow r^2-13r+36=0$
$r^2-9r-4r+36=0$
$\Rightarrow r(r-9)-4(r-9)=0$
$\Rightarrow (r-4)(r-9)=0$
$\therefore r=4$ or $r=9$
$\because r\le n\Rightarrow r\le 5$ and $r\le 6$
$\therefore r=9$ is not possible.
So, $r=4$.