The question has been discussed before here.
The answer one of the respondents has given is $(m-1)! \cdot n! \cdot C(m,n)$.
I found the question solved in here.
Where the author has given the answer as (for no $2$ men sitting together):
$$m! \cdot n! \cdot [C(n-1,m-1) + C(n,m)]$$
I could follow his argument but could not understand why he didn't use $(n-1)!$ instead of $n!$ since we have circular permutations.
Here is the soln in the book. Solution
