Question

Find the value of $n$ such that
${}^n{P}_5=42{}^n{P}_3,n>4$

Answer 1

${}^n{P}_5=42{}^n{P}_3,n>4$

$\Rightarrow \frac{n!}{(n-5)!}=\frac{42\cdot n!}{(n-3)!}\frac{1}{(n-5)!}=42\cdot \frac{1}{(n-3)!}$

$\Rightarrow \frac{1}{(n-5)!}=\frac{42}{(n-3)(n-4)(n-5)!}\therefore 1=\frac{42}{n^2-3n+4n+12}⟹n^2-7n-30=0$

By solving the above equation

$n(n-10)+3(n-10)=0$ i.e, $n=10$ or $-3$

$n$ cannot be negative,

$\therefore n=10$