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

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- $n$ 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 = $^6P_5.5!=6!.5!$

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