# Arranging 20 men and 20 women in a circle while none of the men should be next to a men

I have a simple question in combinatorics:

I want to arrange 20 men and 20 women in a circle while none of the men should stand next to another men.

What I did is this:

I'll stick the first men as a pivot, and arrange the other 19 while leaving spaces for the women to join in between 2 men.

That means: $19!$ permutations for the men.

Now there are 21 places to place the women, so we can arrange them in $20! \cdot 2$ (I multiplied by 2 because there is one more place we can occupy with a women, so if I want to use it as well it's a total different permutation)

So in total: $19! \cdot 20! \cdot 2$

I'm not quite sure about my solution, can anyone please verify if it's true? Anyway, I'd be glad to get some feedback on how I should solve this problem better.

Thanks!

You took care of the circular symmetry by seating a particular man (the King) in a specific chair (the throne). There are now $19!$ ways to arrange the men in alternate seats.
That leaves $20$ "gaps" that can be filled by the women in $20!$ ways, for a total of $19!20!$ ways.