Question

If \( m_1 \) and \( m_2 \) satisfy the relation \( {}^{m+5}P_{m+1} = \frac{11}{2}(m-1)({}^{m+3}P_m) \), then \( m_1 + m_2 \) is

• A. \(10\)
• B. \(9\)
• C. \(13\)
• D. \(17\)
• E. \(15\)

Hint:

Reduce permutations using factorial expansion before solving equations.

Answer 1

The Correct Option is C

Concept: Use: \[ {}^nP_r = \frac{n!}{(n-r)!} \]

Step 1: Expand LHS
\[ {}^{m+5}P_{m+1} = \frac{(m+5)!}{4!} \]

Step 2: Expand RHS
\[ {}^{m+3}P_m = \frac{(m+3)!}{3!} \]

Step 3: Substitute
\[ \frac{(m+5)!}{24} = \frac{11}{2}(m-1)\cdot \frac{(m+3)!}{6} \]

Step 4: Simplify factorials
\[ (m+5)! = (m+5)(m+4)(m+3)! \] Substitute: \[ \frac{(m+5)(m+4)(m+3)!}{24} = \frac{11}{12}(m-1)(m+3)! \] Cancel \((m+3)!\) \[ \frac{(m+5)(m+4)}{24} = \frac{11}{12}(m-1) \]

Step 5: Multiply
\[ (m+5)(m+4) = 22(m-1) \] \[ m^2 + 9m + 20 = 22m -22 \] \[ m^2 -13m + 42 = 0 \] \[ (m-6)(m-7)=0 \] \[ m_1 + m_2 = 6+7 = 13 \] \[ \boxed{13} \]