No two men together...

First seat all the males. This can be done in $\displaystyle 4!=24$ ways. Now, observe that there are 3 gaps between the 4 men (excluding the empty gaps at the left and right side of the leftmost and the rightmost men). So, we need to fill these with 3 women out of the 4. This again can be done in $\displaystyle 4!=24$ ways. Finally, the last woman can now be place in any of the 5 remaining gaps (now, including the before excluded empty gaps).

So, the final answer is $\displaystyle 4!^2\cdot5=2880$

There are 5 patterns:

• mwmwmwmw
• wmwmwmwm
• mwwmwmwm
• mwmwwmwm
• mwmwmwwm

And there are $4! = 24$ ways each sex can arrange themselves within each pattern.

So there are $5\cdot24\cdot24 = 2880$ ways all together.