6 men and 6 women are seated around a table. What is the probability t

Let us first set the restrictions. Let the woman be P and the 2 men be A and B.

There are 12 seats. Let P occupy any of the seats. This can be done in 1 way (As all seats are identical)

The seats to the left and the right cannot be occupied by A and B. They therefore have 9 other seats to choose from.

They can choose 2 seats out of 9 in 9C2 = 36 ways and can arrange themselves in 2! ways (AB or BA)

Finally we have 9 people remaining who can arrange themselves in 9! ways.

Total Favorable outcomes = 36 * 2 * 9!

Total possible outcomes = (12 - 1)! = 11!


The required probability = \(\frac{72 \space * \space 9!}{11!} = \frac{72 \space * \space 9!}{11 \space * \space 10 * \space 9!} = \frac{36}{55}\)


Option D

Arun Kumar