Question

10 men and 6 women are to be seated in a row so that no two women sit together. The number of ways they can be seated is

• A. $11 \cdot 10!$
• B. $\dfrac{11!}{6!5!}$
• C. $\dfrac{10! \, 9!}{5!}$
• D. $\dfrac{11! \, 10!}{5!}$

Hint:

For problems where no two people of a group sit together, first seat the larger group and then place the others in the gaps.

Answer 1

The Correct Option is D

Step 1: First arrange the 10 men in a row. They can be arranged in: \[ 10! \text{ ways} \]
Step 2: These 10 men create 11 gaps (including the two ends) where women can be seated: \[ _ M _ M _ M _ \cdots M _ \]
Step 3: To ensure that no two women sit together, choose 6 gaps out of 11 to place the women: \[ \binom{11}{6} \text{ ways} \]
Step 4: Arrange the 6 women in the selected gaps: \[ 6! \text{ ways} \]
Step 5: Multiply all possible arrangements: \[ 10! \times \binom{11}{6} \times 6! \]
Step 6: Simplify: \[ 10! \times \frac{11!}{6!5!} \times 6! = \frac{11! \, 10!}{5!} \]