Question

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

Answer 1

${}^5{P}_r=\frac{5!}{(5-r)!}$

and ${}^6{P}_{r-1}=\frac{6!}{(6-(r-1))!}=\frac{6!}{(7-r)!}$

Given:${}^5{P}_r=2^6{P}_{r-1}$

$\frac{5!}{(5-r)!}=2\times \frac{6!}{(7-r)!}$

$\frac{(7-r)!}{(5-r)!}=2\times \frac{6!}{5!}$

$\frac{(7-r)(6-r)(5-r)!}{(5-r)!}=2\times 6$

$(7-r)(6-r)=12$

$\Rightarrow r^2-13r+30=0$

$\Rightarrow (r-3)(r-10)=0$

Hence $r=3,10$

But $r<n$

So,$r<5$ and $r<6$

$\therefore r=10$ is not possible.

So,$r=3$