Question

Find the value of $n$ such that
$\left(i\right){}^n{P}_5=42{}^n{P}_3,n>4$
$\left(ii\right)\frac{{}^n{P}_4}{{}^{n-1}{P}_4}=\frac{5}{3},n>4$

Answer 1

$\left(1\right)$ Step $1:$
Calculating ${}^n{P}_5$

${}^n{P}_5=\frac{n!}{(n-5)!}$
${}^n{P}_5=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)!}{(n-5)!}$
${}^n{P}_5=n(n-1)(n-2)(n-3)(n-4)...\left(i\right)$

Step $2:$
Calculating $42{}^n{P}_3$

$42{}^n{P}_3=\frac{42n!}{(n-3)!}$
$=\frac{42n(n-1)(n-2)(n-3)!}{(n-3)!}$
$=42n(n-1)(n-2)...\left(ii\right)$

Step $3:$ By comparing $\left(i\right)$ and $\left(ii\right)$
$n(n-1)(n-2)(n-3)(n-4)=42n(n-1)(n-2)$
$\Rightarrow (n-3)(n-4)=42$
$\Rightarrow n^2-3n-4n+12-42=0$
$\Rightarrow n^2-7n-30=0$
$\Rightarrow n^2-10n+3n-30=0$
$\Rightarrow n(n-10)+3(n-10)=0$
$\Rightarrow (n-10)(n+3)=0$
$\Rightarrow n=10$ or $n=-3$
But here, $n>4$ so $n=-3$ not possible.
$\therefore n=10$

$\left(ii\right)$ Step $1:$ Calculating ${}^n{P}_4$
${}^n{P}_4=\frac{n!}{(n-4)!}$
$=\frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}$
$=n(n-1)(n-2)(n-3)...\left(i\right)$

Step $2$
Calculating ${}^{n-1}{P}_4$
${}^{n-1}{P}_4=\frac{(n-1)!}{(n-1-4)!}$
$=\frac{(n-1)!}{(n-5)!}$
${}^{n-1}{P}_4=\frac{(n-1)(n-2)(n-3)(n-4)(n-5)!}{(n-5)!}$
${}^{n-1}{P}_4=(n-1)(n-2)(n-3)(n-4)...\left(ii\right)$

Step $3$ On solving

$\frac{{}^n{P}_4}{{}^{n-1}{P}_4}=\frac{5}{3}$
$3{}^n{P}_4=5^{n-1}{P}_4$
$3n(n-1)(n-2)(n-3)=5(n-1)(n-2)(n-3)(n-4)$
$\Rightarrow 3n=5(n-4)$
$\Rightarrow 3n=5n-20$
$\Rightarrow 2n=20$
$\Rightarrow n=10$