Question

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

• A. $5$
• B. $6$
• C. $7$
• D. $4$

Answer 1

Answer: (B), (C)

$7$
$6$

$\frac{(n+5)!}{4!}=\frac{11(n-1)(n+3)!}{2.3!}$
$\frac{(n+5)(n+4)}{2}=\frac{11(n-1)}{1}$
$(n^2+9n+20)=22(n-1)$
$(n^2+9n+20)=(22n-22)$
$n^2-13n+42=0$
$(n-6)(n-7)=0$
$=>n=6$ or $7$