Question

If ${_{}^{n}P}_{5} = 20.\mspace{6mu}{_{}^{n}P}_{3}$, then $n =$

• A. 4
• B. 8
• C. 6
• D. 7

Answer 1

Answer: (B)

8

Answer 2

$\frac{n\mspace{6mu}!}{(n - 5)\mspace{6mu}!} \times \frac{(n - 3)\mspace{6mu}!}{n\mspace{6mu}!}$=20

$\Rightarrow (n - 3)(n - 4) = 20 \Rightarrow n = - 1,\mspace{6mu} 8$

But $- 1$ is not exceptable.