Question

Find n if ${}^n{P}_4={18}^{n-1}{P}_2$

• A. 4
• B. 8
• C. 6
• D. 12

Answer 1

The correct option is C 6

${}^n{P}_4={18}^{n-1}{P}_2$
\Rightarrow \frac{n!}{(n-4)!} = 18. \frac{(n-1)!}{(n-1-2)!}\\
\Rightarrow \frac{n!}{(n-4)!} = 18. \frac{(n-1)!}{(n-3)!}\\
\Rightarrow \frac{n(n-1)!}{(n-4)!} = 18. \frac{(n-1)!}{(n-3)(n-4)!}\\
\therefore n = \frac{18}{n-3}\\
i.e.,~n^2 - 3n - 18 = 0 \Rightarrow (n-6)(n+3)=0\\
\Rightarrow n = 6, -3\)
But n cannot be negative
$\therefore n=6$