We have three tasks, here
1. Choose any 2 of 5 men to sit at the beginning of the row.....and arrange these in any order = 5P2 = 20 possible arrangements
2. Choose any 2 of the 3 remaining men to sit at the end of the row and arrange them in any order = 3P2 = 6 possible arrangements
3. Arrange the 5 people left in the middle in any order = 5! = 120 possible arrangements
So ......20 * 120 * 6 = 120^2 = 14,400 possible arrangements
[ Your problem doesn't specify how the people in the middle must be seated.....we could have three men sitting together at either the start of the row or at the end of the row in my scenario......however....two men are still sitting at each end]
However.....the problem changes slightly if we require that only two men can sit at both ends....
In that case......(1) and (2) above are still good
However......the remaining man can only be seated in 3 ways [positions 4-6] and the other 4 women can be seated in 4! ways = 24
So....with this restriction....the number of arrangements is :
20 * 6 * 3 * 24 = 8640 arrangements
