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.
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
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