Question

(i) If ${}^n{P}_4{:}^n{P}_5=1:2$ find n.

(ii) If ${}^{n-1}{P}_3{:}^{n+1}=5:12,$ find n.

Answer 1

$(i{)}^n{P}_4{:}^n{P}_5=1:2$

$\frac{\frac{n!}{(n-4)!}}{\frac{n!}{(n-5)!}}=\frac{1}{2}$

$\frac{n!}{(n-4)!}\times \frac{(n-5)!}{n!}=\frac{1}{2}$

$\frac{1}{n-4}=\frac{1}{2}\Rightarrow n=6$

$\left(ii\right)\frac{(n-1)(n-2)(n-3)}{(n+1)n(n-1)}=\frac{5}{12}$

$\Rightarrow 12(n-2)(n-3)=5(n+1)n$

$\Rightarrow 7n^2-65n+72=0\Rightarrow 7n^2-56n-9n+72=0$

$\Rightarrow 7n(n-8)-9(n-8)=0\Rightarrow (n-8)(7n-9)=0\Rightarrow n=8.$