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 $^6P_5$ ways.
$\therefore$ the required no. of arrangements = $^6P_5.5!=6!.5!$