5 men and 5 women are to be seated in a row of 10 seats. How many ways can the men not sit together and the women not sit together?
If you want to find the number of ways to arrange men and women so that we have no men or women sitting consecutively we get
$$2\cdot5!\cdot5!$$ Since There $5!$ ways to arrange the males skipping a seat in between and $5!$ ways to arrange the females in the remaining $5$ spots. Then there are $2$ ways to start the row, either male or female.
Also if you're trying to find the number of ways to only group men together and women together (MMMMMWWWWW and WWWWWMMMMM) the number is the same, $2\cdot5!\cdot5!$
I’m assuming that you mean that no two men sit next to each other, and no two women sit next to each other. If you mean that the arrangement is not of the form MMMMMWWWWW or WWWWWMMMMM, the answer will be different.
HINT: The men and women must alternate, either WMWMWMWMWM or MWMWMWMWMW. In each of those general arrangements how many ways are there to arrange the men? How many ways are there to arrange the women? How should you combine these two numbers to get the total number of WMWMWMWMWM arrangements?
Let us find the number of ways the men are seated in a clump or the women are seated in a clump or both.
If the men are to be in consecutive seats, then the seats labelled M can be chosen in $6$ ways, for the leftmost one can be any of $1$ to $6$. The same is true if the seats labelled W are to be consecutive. If we find the sum $6+6$, we will be double-counting the ways in which the M seats are together and the W seats are together. So the actual number of ways to choose the seats for men and the seats for women is $10$.
For every way of choosing the seats for men and women, the seating can take place in $5!5!$ ways. So the total number of ways in which the men are not together and the women are not together is $10! -(10)(5!5!)$.