How many ways can 8 people be seated in a row? I am stuck with the following question,


  How many ways can 8 people be seated in a row?
  if there are 4 men and 4 women and no 2 men or women may sit next to each other.


I did it as follows,
As 4 men and 4 women must sit next to each other so we consider each of them as a single unit.
Now we have we 4 people(1 men group, 1 women group, 2 men or women) they can be seated in 4! ways.Now each of the group of men and women can swap places within themselves so we should multiply the answer with 4!*4! 
This makes the total  4!*4!*4! =13824 .
Please help me out with the answer. Are the steps clear and is the answer and the method right?

Thanks
If there is a man on the first seat, there has to be a woman on the second, a man on the third so forth. Alternatively, we could start with a woman, then put a man, then a woman and so forth. In any case, if we decide which gender to put on the first seat, the genders for the others seats are forced upon us. So there are only two ways in which we can divide the eight aligned seats into "man-seats" and "woman-seats" without violating the rule that no two men and no two women may sit next to each other.

Once you have chosen where to seat the men and where the women you can permute the two groups arbitrarily, giving $4!$ possibilities each. So in total the number of possible constellations is
$$
2\cdot 4!\cdot 4!=1152.
$$
Is it that no two men and no two women should be seated together. 
Yep? Then do this..
Permutation of men in 4 places be 4!. This leaves us with 5 gaps(3 in between and 2 on either side) in which we got to arrange 4 women.. P(5,4) i.e. 5! ways.
Finally   5!*4! is the answer!.