In how many different orders can 4 men and 4 women be seated in a row so that no person of the same sex will be together?
I've read that the equation is $n!(n-1)!$
Then I read another one that says it's $2 \cdot 4! \cdot 4!$
Which one is correct, and when does the other equation apply?
There are only two options for the sequence of genders, MFMFMFMF or FMFMFMFM. For each of these $2$ choices you can distribute the men in $4!$ ways and the women in $4!$ ways, for a total of $2\cdot 4!\cdot 4!$ possibilities.
Well, if they are seated in a row, then the first in the row can be man or woman. Once we know who is the first, the positions for men and women are fixed (MWMW...). So we will remember to multiply by 2 at the end.
You have 4 places for men and 4 for women, the possibilities for each are $4!$, so the total number of arrangements is $4! \cdot 4!$. Then we multiply by our $2$ and get the solution $2\cdot 4! \cdot 4!$.
As Henning Makholm has explained, the answer to this question is $2 \cdot 4! \cdot 4!$.
The formula $n!(n + 1)!$ is the answer to the question:
In how many ways can $n$ men and $n$ women be seated in a row so that no two men sit in consecutive seats?
To see why, consider your example of four men and four women. We can arrange the women in $4!$ ways. To ensure that no two men sit in consecutive seats, we must choose four of the five spaces indicated by a square in which to place a man, then arrange the men in those spaces.
$$\square W \square W \square W \square W \square$$
Hence, the men can be arranged in $$P(5, 4) = \binom{5}{4}4! = \frac{5!}{4!1!} \cdot 4! = 5!$$ ways. Thus, in this example, the number of seating arrangements in which no two of the four men sit in consecutive seats is $4!5!$.
In the general case, we can arrange the $n$ women in $n!$ ways, leaving $n + 1$ spaces in which to place the men, namely the $n - 1$ spaces between successive women and the two ends of the row. To ensure no two men are adjacent, we choose $n$ of these $n + 1$ spaces in which to place the men, then arrange the men in these selected spaces, which can be done in $$\binom{n + 1}{n} = \frac{(n + 1)!}{n!1!} \cdot n! = (n + 1)!$$
ways, which gives $$n!(n + 1)!$$ permissible seating arrangements.