Question

If ${}^{12}{P}_r{=}^{11}{P}_6+{6.}^{11}{P}_5$, then the value of $r$ is

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

Answer: (B)

$6$

Given equation is ${{}^{12}P}_r={{}^{11}P}_6+6\times {{}^{11}P}_5$

We know that ${{}^{n}P}_r={{}^{n-1}P}_r+r.{{}^{n-1}P}_{r-1}$

Thus ${{}^{12}P}_6={{}^{12-1}P}_6+6\times {{}^{12-1}P}_{6-1}={{}^{11}P}_6+6\times {{}^{11}P}_5$

Thus ${{}^{12}P}_r={{}^{12}P}_6\Rightarrow r=6$