Question

${}^{15}{P}_8=A+8\cdot {}^{14}{P}_7\Rightarrow A=$

• A. ${}^{14}{P}_6$
• B. ${}^{14}{P}_8$
• C. ${}^{15}{P}_7$
• D. ${}^{16}{P}_9$

Answer 1

Answer: (B)

Given, ${}^{15}{P}_8=A+8\cdot {}^{14}{P}_7$
$\Rightarrow \frac{15!}{7!}=A+8\cdot \frac{14!}{7!}$
$\Rightarrow A=\frac{14!}{7!}(15-8)=\frac{14!}{7!}(7)$
$=\frac{14!}{6!}={}^{14}{P}_8$