In how many ways can they sit so that no 2 men are sitting next to each other?

Question

There 14 women and 8 men and they want to sit on a circular table. In how many ways can they sit so that no 2 men are sitting next to each other?

There are $(14-1)!=13!$ ways in which women can sit.

Then there are $\binom{14}{8}$ so that men can sit, or not?

So do we get the answer by multiplying these two ?

Answer 1

First of all select position for men. Men can sit in $\binom{14}{8}$ position .

Then both women and men can permute in $(14-1)!$ and $(8)!$ ways respectively.

Final answer=$$\binom{14}{8} \times 13! \times 8!$$

Answer 2

I preassume that the persons are distinguisable and the seats are not.

Place one of the men at the table. Starting at his left hand there comes a sequence of the form:$$.m.m.m.m.m.m.m.$$where every dot stands for at least one woman and every $m$ for one man.

This comes to finding the number of sums $w_1+\cdots+w_8=14$ where the $w_i$ are positive integers, or equivalently to finding the number of sums $v_1+\cdots+v_8=6$ where the $v_i$ are nonnegative integers.

Without distinghuishing persons and applying inclusion/exclusion we find that this can be achieved on $\binom{13}7$ ways. There are $7!$ orders for men and $14!$ for women, so the final answer is:$$\binom{13}77!14!=\frac{13!14!}{6!}$$If also seats are distinguisable then this must be multiplied with factor $22$ corresponding with the number of seats available for the man that was placed at the table as first.