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

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

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

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

$\Rightarrow 1=\frac{12}{(7-r)(6-r)}$

$\Rightarrow r^2-13r+42=12$

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

$\Rightarrow r^2-10r-3r+30=0$

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

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

$\Rightarrow r=10orr=3$

Now $r=10$ is not possible because $r>n.$

Thus $r=3.$

(ii) ${}^{5}Pr{=}^6{P}_{r-1}$

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

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

$\Rightarrow 1=\frac{6}{(7-r)(6-r)}$

$\Rightarrow r^2-13r+42=6$

$\Rightarrow r^2-13r+36=0$

$\Rightarrow r^2-9r-4r+36=0$

$\Rightarrow r(r-9)-4(r-9)=0$

$\Rightarrow (r-9)(r-4)=0$

$\Rightarrow r=9orr=4$

Now $r=9$ is not possible because $r>n.$

Thus $r=4.$