Question

If P(5,r) = P (6,r-1), find r.

Answer 1

We have,
P(5,r) = P(6,r-1)
$\Rightarrow \frac{5!}{(5-r)!}=\frac{6!}{[6-(r-1\left)\right]!}$
$[\because n_{P_r}=\frac{n!}{(n-r)!}]$
$\Rightarrow \frac{1}{(5-r)!}=\frac{6!}{[7-r]!}$
$\Rightarrow \frac{1}{(5-r)!}=\frac{6}{(7-r)\times (7-r-1)(7-r-2)!}$
$\Rightarrow \frac{1}{(5-r)!}=\frac{6}{(7-r)\times (6-r)(5-r)!}$
$\Rightarrow 1=\frac{6}{(7-r)\times (6-r)}$
$\Rightarrow (6-r)\times (7-r)=6$
$\Rightarrow 42-6r-7r+r^2=6$
$\Rightarrow r^2-12r+42-6=0$
$\Rightarrow r^2-12r+42-6=0$
$\Rightarrow r^2-13r+36=0$
$\Rightarrow r^2-9r-4r+36=0$
$\Rightarrow r^2-9r-4r+36=0$
$\Rightarrow r(r-9)-4(r-9)=0$
$\Rightarrow (r-9)(r-4)=0$
$\Rightarrow r=4\left[\begin{array}{c}\because r\le n\\ \therefore r-9\ne 0\end{array}\right]$
Hence, r= 4