# The no. of ways in which 6 men and 5 women can sit at a round table so that no two women sit together is?

$\begin{array}{1 1} 30 \\ 5! 4! \\ 7 !5! \\ 6!5! \end{array}$

Toolbox:
• $n$ different things can be arranged in $(n-1)!$ ways in a circle.
• $^nP_r=\large\frac{n!}{(n-r)!}$
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 = $^6P_5.5!=6!.5!$