Question

If ${}^{3n}{P}_{n-1}:{}^{3n+1}{P}_n=1:4,$ then $n$ is equal to

• A. $4$
• B. $6$
• C. $1$
• D. $3$

Answer 1

Answer: (C)

$1$

$\Rightarrow \frac{{}^{3n}{P}_{n-1}}{{}^{3n+1}{P}_n}=\frac{1}{4}\Rightarrow \frac{\frac{\left(3n\right)!}{(3n-n+1)!}}{\frac{(3n+1)!}{(3n-n+1)!}}=\frac{1}{4}\Rightarrow \frac{1}{3n+1}=\frac{1}{4}\Rightarrow 3n+1=4\Rightarrow 3n=3$

Hence the correct answer is $1$