Question

Find $r$, if $5{}^4{P}_r=6{}^5{P}_{r-1}$

Answer 1

$5{}^4{P}_r=6{}^5{P}_{r-1}$
$\Rightarrow 5\times \frac{4!}{(4-r)!}=6\times \frac{5!}{(5-(r-1\left)\right)!}$
$\Rightarrow 5\times \frac{4!}{(4-r)!}=6\times \frac{5!}{(5-r+1)!}$
$\Rightarrow 5\times \frac{4!}{(4-r)!}=6\times \frac{5!}{(6-r)!}$
$\Rightarrow \frac{(6-r)!}{(4-r)!}=\frac{6\times 5!}{5\times 4!}$
$\Rightarrow \frac{(6-r)(5-r)(4-r)!}{(4-r)!}=\frac{6\times 5\times 4!}{5\times 4!}$
$\Rightarrow (6-r)(5-r)=6$
$\Rightarrow 30-6r-5r+r^2=6$
$\Rightarrow r^2-11r+24=0$
$\Rightarrow r^2-8r-3r+24=0$
$\Rightarrow r(r-8)-3(r-8)=0$
$\Rightarrow r-8=0$ or $r-3=0$
$\Rightarrow r=8$ or $r=3$
$\because r<n$
$\Rightarrow \therefore r=8\to$ Not possible
So, $r=3$