Verification on how I'm approaching this permutation/counting problem 
Let $f \geq 4$ and $m \geq 4$ be integers. There are $f$ female students and $m$ male students that are eligible to be members on the council.
Determine the number of way to choose eight members for the council
  out of these $f + m$ students, such that the number of female members is
  equal to the number of male members.

This is a question out of my textbook's practice problems which has no answer keys. It's hard to know if I'm doing things right without being able to confirm my answer. Was wondering if someone can help me out with this.
We have $8$ possible members, if there are equal male and female then there has to be $4$ male members and $4$ female members. Would the answer to this problem simply be a permutation problem. $8$ choose $4$?
$$p = \frac{8!}{4!4!}$$
 A: Choose $4$ male students from $m$ available: ${\large{\binom{m}{4}}}$ choices.

Choose $4$ female students from $f$ available: ${\large{\binom{f}{4}}}$ choices.

The total number of possible committees is the product: ${\large{\binom{m}{4}}}{\large{\binom{f}{4}}}$.

Explanation: For each of the ${\large{\binom{m}{4}}}$ choices for the $4$ male students, there are ${\large{\binom{f}{4}}}$ choices for the $4$ female students, so the product rule can be applied.  
A: You are right that if we were to choose $8$ students such that the number of female members is equal to the number of male members, then we have to choose $4$ male members and $4$ female members.
Let us break the whole process into two procedures.
Firstly, there are $\dbinom{m}{4}$ possibilities of choosing $4$ male members from $m$ male students.
Secondly, there are $\dbinom{f}{4}$ possibilities of choosing $4$ female members from $f$ female students.
Now, using the product rule, there are $\dbinom{m}{4} \times \dbinom{f}{4}$ ways of choosing $4$ male members from $m$ male students and choosing $4$ female members from $f$ female students.
Note that, we can first choose $4$ female members from $f$ female students, followed by choosing $4$ male members from $m$ male students. The entire process is the same.
You can verify it by noting that $\dbinom{f}{4} \times \dbinom{m}{4}$ = $\dbinom{m}{4} \times \dbinom{f}{4}$.
