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 $\binom{8}{4}\times \binom{7}{3}=2450$.
 A: To solve this, we find the number of ways of choosing the women, and multiply that by the number of ways of choosing the men.
So we begin by counting the $4$ women out of the $8$ in total. That's $\binom 84$ ways to count the women.
Call the feuding men $A$ and $B$. Then one and only one of the following can be true for any team configuration:
(i) A is on the team
(ii) B is on the team
(iii) Neither are on the team.
Case (i): We choose $2$ of the remaining $5$ (we guarantee A and exclude B).
Case (ii): We choose $2$ of the remaining $5$ (we guarantee B and exclude A).
Case (iii): We choose $3$ of the remaining $5$ (we exclude A and B).
This yields $\binom52 + \binom52 + \binom53$ combinations for the men.
Now we multiply them together to acquire $$\binom 84 \Bigg(\binom52 + \binom52 + \binom53 \Bigg) = 2100$$

Alternative method. Sans restriction, we can choose $3$ men from the $7$ we have in $\binom 73$ ways. Now we need to subtract the cases that we shouldn't have counted. The cases that we shouldn't have counted were those in which the who feuding men were together. If they're together, then there are $\binom{7-2}{1} = 5$ ways to choose the last man, which means $5$ of our counts should be excluded. And so we have $$\binom 84 \Bigg(\binom73 - 5 \Bigg) = 2100$$
A: It is important to understand why the book answer is incorrect.
The first term for counting men overcounts. $\dbinom62$ leaves out $A$, say, from contention, but $B$ still remains in contention !
The correct way for the 1st term is to remove both from contention, and choose one of the two, thus
$\dbinom84\left[\dbinom52\dbinom21 + \dbinom53\right] = 2100$   
A: 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.
