The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

Six men can be sited on a round table in 5 ! ways.

Now, in six places between six men, five women can be seated

in ${ }^6{C}_5\times 5!$ ways.

Hence, required number of seating arrangements

$=5!{\times }^6{C}_5\times 5!=6!\times 5!$.