Number of teams There are $8$ women and $7$ men, from which we must create a team of $4$ women and $3$ men. Two men doesn't like each other, so they don't want to be in one team.
I tried number of teams (without the specific two deviant men): $\binom{8}{4}\times \binom{7}{3}$ but it's not correct.  
The given solution is: $\binom{6}{2}\binom{8}{4}\cdot2+\binom{5}{3}\cdot \binom{8}{4}=2800$, which is more than $\binom{8}{4}\times \binom{7}{3}=2450$.

Some thoughts:
1) No matter what, the number of women on the team can be computed as $\binom{8}{4}$
2) $\binom{8}{4}\times \binom{7}{3} = 2450$. $\binom{8}{4}\times \binom{6}{2}+\binom{8}{4}\times \binom{5}{3} = 1750$, not $2800$.
3) Count the teams containing neither of the special men. Then add the count of teams containing the first special man. Next add the count of teams containing the second special man. Finally, multiply that result by the (independent) count  of the women. Once you see how to break up the count, the numbers work out easily.
