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

Look , we have 8 people , 4 of them are men , 4 of them are women , we need to put every man and woman together , it means we will have 4 groups of men and women sitting together. [Group1,Group2,Group3,Group4] , to get  Group1 we have 4 possibilities of men , and 4 possibilities of women , then to get Group2 we left with 3 men and 3 women , then in the same manner we got  : 4*4*3*3*2*2*1*1 =576 (4!*4!) , so finally we left only with combinations of the groups in the array [Group1,Group2,Group3,Group4], as I got there 4! # of combinations this groups .
So I guess this is the answer : 4!*4!*4!
