Question

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

Answer 1

(i)

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

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

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

$(7-r)(6-r)=2\times 6$

$42-7r-6r+r^2=12$

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

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

$\therefore r=3,10$

(ii)

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

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

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

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

$42-7r-6r+r^2=6$

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

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

$r=4,9$