TITLE: Counting, arranging 8 ppl
QUESTION [6 upvotes]: I should be able to solve this Problem in my sleep but i cant figure out why my solution is not right - i even have thought about the possibility that the textbook solution is wrong but I guess that would be pretty arrogant.
So here is the Question.
In how many ways can 8 people be seated in a row if there are 4 men and 4 women and no 2 men or 2 women should sit next to each other. 
My solution and thoughts are as follows:
There are 4! ways to arrange the women and 4! ways to arrange the men, and the row can start with either a men oder a women so times two 
4!*4!*2=1152
But the textbook solution shows 
5! *4!=2880 and I have no idea how they come up with that and what is wrong with my solution.
I would be very grateful for every hint that could help me understand.

REPLY [2 votes]: The textbook answer is incorrect.
Given condition is that no two men can sit next to each other, and the possible way to be seated that way is 
                            M_M_M_M_ 

These $4$ men can be seated in $4!=24$ ways
Now, the remaining empty seats can be filled by women in $4!=24$ ways.
This can be done in $2$ ways.
Therefore, the number of ways to arrange $8$ people in a row if there are $4$ men and $4$ women and no $2$ men or $2$ women can sit next to each other is $2\times4!\times4!=2\times24\times24=1152$ ways