Question

If ${}^{12}{P}_r{=}^{11}{P}_6+{6.}^{11}{P}_5$ then r is equal to

• A. 6
• B. 5
• C. 7
• D. none of these

Answer 1

The correct option is A 6

${}^{11}{P}_6+6.{}^{11}{P}_5=\frac{11!}{(11-6)!}+6.\frac{11!}{(11-5)!}$

$=\frac{11!}{5!}+6.\frac{11!}{6!}$

$=\frac{11!}{5!}+\frac{11!}{5!}=2\left(\frac{11!}{5!}\right)$

$=2\times \frac{6}{6}\times \frac{11!}{5!}$

$=\frac{12\times (11!)}{6\times (5!)}$

$=\frac{12!}{6!}$

$=\frac{12!}{(12-6)!}$

$={}^{12}{P}_6$

$\therefore {}^{11}{P}_6+6{}^{11}{P}_5={}^{12}{P}_6$.