Question

Find the no. of ways in which $6$ men and $5$ women can sit at a round table so that no two women sit together?

Answer 1

Since no conditions for men,

$6$ men can be arranged in $5!$ ways in a round table.

In between each man a woman is to be placed since no two women can be together.

There are $6$ gaps between two men and $5$ women are to be placed.

This can be done in ${}^6{P}_5$ ways.

$\therefore$ the required number of arrangements ${=}^6{P}_5.5!=6!.5!=86,400‬$ways