Question

If ${}^{n+5}{P}_{n+1}=\frac{11(n-1)}{2}{.}^{n+3}{P}_n$ then the value of n is

• A. 7
• B. 8
• C. 6
• D. 5

Answer 1

Answer: (A), (C)

7
6

Applying the given condition, we get
$\frac{(n+5)!}{4!}=\frac{11(n-1)}{2}\frac{(n+3)!}{3!}$
$\frac{(n+4)(n+5)}{4}=\frac{11(n-1)}{2}$
$(n+4)(n+5)=22(n-1)$
$n^2+9n+20=22n-22$
$n^2-13n+42=0$
$(n-7)(n-6)=0$
$n=7$ $n=6$