Question

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

• A. $10$
• B. $15$
• C. $12$
• D. $20$

Answer 1

Answer: (A)

We have ${}^n{P}_5=42^n{P}_3$
$\Rightarrow n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)=42n\left(n-1\right)\left(n-2\right)$
$\Rightarrow \left(n-3\right)\left(n-4\right)=42$
[Since $n>4$, so $n\left(n-1\right)\left(n-2\right)\ne 0$]
$\Rightarrow n^2-7n-30=0$
$\Rightarrow n^2-10n+3n-30=0$
$\Rightarrow n=10$ or $n=-3$
As n cannot be negative, so $n=10$