Question

Find the value of $n$ such that ${}_n{P}_5=42{(}_n{P}_3):n>4$.

Answer 1

It is given that ${}^n{P}_5=42{}^n{P}_3$, then,

$\frac{n!}{\left(n-5\right)!}=42\frac{n!}{\left(n-3\right)!}$

$\frac{n!}{\left(n-5\right)!}=42\frac{n!}{\left(n-3\right)\left(n-4\right)\left(n-5\right)!}$

$\left(n-3\right)\left(n-4\right)=42$

$n^2-7n+12-42=0$

$n^2-7n-30=0$

$n^2-10n+3n-30=0$

$n\left(n-10\right)+3\left(n-10\right)=0$

$\left(n+3\right)\left(n-10\right)=0$

$n=-3,n=10$

Since, it is given that $n>4$, then only value of $n$ is $n=10$.