Weird arrangement

Two basic arrangements: MWMWMW or WMWMWM.

Within each there are 6 possible positions for the men and 6 for the women, independent of each other.

Total is then $6\cdot6\cdot2=\boxed{72}$ .

The main point of the question is that no men and women can sit adjacent to each other. So if the first woman sit on the first chair then the other two woman must occupy the other odd postions and the three men are sit on the even positoned chairs. Now assuming that the first woman on the first chair, second on the third chair and third on the fifth chair the three men can sit on the other chairs in P(3,3) or 3! ways. The 3 womens can sit on the odd positioned chairs also in 3! ways and for every arrangements of women there are 3! ways of arrangement of men. So there are a total of 3! 3! ways. But there are still another arrangement can be done. Now consider the 3 women occupy the even positioned chairs and all men men occupy the odd ones. Now in a similar manner it can be shown that there are 3! 3! ways of arrangement. So there are a total of 3! 3! + 3! 3! or 2*3!^2 or 72 ways.