I have 2 different problems that seem identical:
8 men and 8 women are going to sit at a round table where there are 16 seats. They take their seats randomly. How many ways can the 16 seats be taken so that no 2 women are sitting next to each other?
I solved this by doing $$2 * 8!8!$$
The method I used was that $8*8*7*7*6*6*5*5*4*4*3*3*2*2*1*1= 8!8!$ is all the ways you can sit these 8 women alternating them with the 8 men.
I then multiplied by two because you have 2 possibilities: the women can be sitting at the odd numbered seats, or at the even numbered ones.
Then I have this other problem:
In how many ways can a party of 4 men and 4 women be seated at a circular table so that no two women are adjacent?
(I took the problem from here: http://www.imsc.res.in/~kamalakshya/cupboard/comb_mag.pdf)
And the solution, according to the website, is:
Answer: The 4 men can be seated at the circular table such that there is a vacant seat between every pair of men in (4-1)! =3! Ways. Now 4 vacant seats can be occupied by 4 women in 4! Ways. Hence the required number of seating arrangements = 3!4! = 144
However, I can't seem to use the method from the 1st problem on this one. All the possible combinations of men and women, sitting alternated with men, would be $4!4!$ Then, multiplying by 2 the solution would be $2 * 4!4! = 1152$
If I use the method from the answer of the 2nd problem with the 1st one, I get $7!8!$, which is different from $2 * 8!8!$
My question is, how can this be? Is there any difference between these 2 problems that I am not seeing?
EDIT: I believe the 1st solution is correct. Check here Probability of men and women sitting at a table alternately