a group consists of 21 members, 11 of which are women 10 are men . A new committee of 8 members is to be selected, with precisely 4 women. the problem I'm facing is in regards to when I have a restriction in how to put together the committee. i have the following restriction.
"Two of the male students in the council strongly dislike each other, and should not be in on the committee together"
The solution i have so far:

It just seems like a way too large a number. Is my method correct?
Please also note that there needs to be precisely 4 women in the group at all time.
 A: Your method is correct, in that $$P = \binom{11}{4}\binom 21 \binom 83 + \binom{11}{4} \binom20 \binom 84$$ is the correct answer. Dividing up the $21$ members of the council into $11$ women, $2$ enemies, and $8$ other men, we can either choose $4$ women, $1$ enemy, and $3$ other men (the first term) or $4$ women, $0$ enemies, and $4$ other men (the second term).
The calculations go wrong in several places; for one, $\binom 83 = \frac{8!}{5!\,3!}$, not $\frac{8!}{2!\,3!}$, and $\binom 84 = \frac{8!}{4!\,4!}$, not $\frac{8!}{4!\,3!}$. Even after that, $2 \cdot 28 + 1 \cdot 280$ is nowhere close to $7000$.
The correct calculation is
$$
   P = 330 \cdot 2 \cdot 56 + 330 \cdot 1 \cdot 70 = 330 \cdot 182 = 60060.
$$
A: Misha Lavrov's answer directly dissects the OP's (i.e. original poster's) work, as requested.  An alternative approach is
$$\binom{11}{4} \times \left[\binom{10}{4} - \binom{8}{2}\right]. \tag1 $$
In the 2nd factor of (1) above, the 2nd term represents the deduction of the number of ways of violating the constraint by having both of the pertinent two males be on the committee.
