Question

5 men and 4 women are seated in a row. If the number of arrangements in which one particular man and one particular woman are together is $ \alpha $, and the number of arrangements in which they are not together is $ \beta $, then $ \frac{\alpha}{\beta} = $

• A. \( \frac{2}{7} \)
• B. \( \frac{2}{9} \)
• C. \( \frac{4}{5} \)
• D. \( \frac{7}{2} \)

Hint:

When dealing with “together” and “not together” arrangements, treat the pair as one block and subtract from total permutations.

Answer 1

The Correct Option is A

Total people = 9 Total arrangements = \( 9! \) Case 1: Man and woman together (treat them as a pair): They can sit in 2 ways (M-W or W-M) Remaining 7 people: 7! So total = \( 2 \times 8! \) So \( \alpha = 2 \times 8! \) Case 2: Not together = total - together \[ \beta = 9! - 2 \times 8! \] Now: \[ \frac{\alpha}{\beta} = \frac{2 \times 8!}{9! - 2 \times 8!} = \frac{2}{9 - 2} = \frac{2}{7} \]