Question

If ${}^{m+5}{P}_{m+1}=\frac{11}{2}(m-1)\left({}^{m+3}{P}_m\right)$ then $m_1+m_2$ is equal to

• A. $10$
• B. $9$
• C. $13$
• D. $17$
• E. $15$

Answer 1

Answer: (C)

$13$

${}^{m+5}{P}_{m+1}=\frac{11}{2}(m-1)\left({}^{m+3}{P}_m\right)$
$\Rightarrow \frac{(m+5)!}{(m+5-m-1)!}=\frac{11}{2}(m-1)\cdot \frac{(m+3)}{(m+3-m)!}$
$\Rightarrow \frac{(m+5)!}{4!}=\frac{11}{2}(m-1)\frac{(m+3)!}{3!}$
$\Rightarrow \frac{(m+5)(m+4)(m+3)!}{4\times 3!}=\frac{11}{2}\frac{(m-1)(m+3)!}{3!}$
$\Rightarrow (m+5)(m+4)=22(m-1)$
$\Rightarrow m^2-13m+42=0$
Since, $m_1$ and $m_2$ satisfy above equation
So, $m_1+m_2=\frac{-(-13)}{1}=13$