A committee of 15 -- 9 women and 6 men -- is to be seated at a circular table (with 15 seats). In how many ways can the seats be assigned so that no two men are seated next to each other?
I am thinking like if we had 5 men 5 women we have 5!5! ways. Similarly we can say lets first put 6 men. Then to divide them put 6 women. Then still we have 3 women. Lets also but remaning 3 women by placing between existing people. 6! . p(9,6) . 12. 13. 14
Select any woman. There are then 8! ways of placing the other women clockwise from her. There are then 9 potential gaps between adjacent women. At most one man can sit in any of these potential gaps. We select 6 of them in ${9\choose6}$ ways. There are then 6! ways of placing the men in those gaps. So the men can be seated in $9!/3!$ ways. So far we have got $$8!\ 9!/3!=6720\cdot9!=2438553600\quad(*)$$ At this point you have to decide whether moving everyone a seat to the left gives you a different arrangement, or whether you only care about the order. If it matters who sits in a particular seat, then we need to multiply the number so far by 15 to get $$\frac{5}{2}\cdot8!\cdot9!=36578304000$$ If you only care about the order, then $(*)$ is the answer.