Question

If ${}^{k+5}{P}_{k+1}$ = $\frac{11(k-1)}{2}$ ${}^{k+3}{P}_k$, then the values of k are

• A. 7 and 11
• B. 6 and 7
• C. 2 and 11
• D. 2 and 6

Answer 1

The correct option is B

6 and 7

${}^{k+5}{P}_{k+1}=\frac{11(k-1)}{2}{}^{k+3}{P}_k$

$\Rightarrow \frac{(k+5)!}{(k-5-k-1)!}=\frac{11(k-1)}{2}\times \frac{(k+3)!}{(k+3-k)!}$

$\Rightarrow \frac{(k+5)!}{4!}=\frac{11(k-1)}{2}\times \frac{(k+3)!}{3!}$

$\Rightarrow \frac{(k+5)!}{(k+3)!}=\frac{11(k-1)}{2}\times \frac{4!}{3!}$

$\Rightarrow (k+5)(k+4)=22(k-1)$

$\Rightarrow k^2+9k+20=22k-22$

$\Rightarrow k^2-13k+42=0$

$\Rightarrow k=6,7$