Question

$5_{P_r}={2.6}_{P_{r-1}}$ find $r$

Answer 1

Given that:

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

$\frac{5!}{(5-r)!}=\frac{6!}{(6-r+1)!}$

$(6-r)(5-r)=6$

$r^2-11r+30=6$

$r^2-8r-3r+24=0$

$(r-8)(r-3)=0$

$r=3,8$

Hence, this is the answer.