Men and Women: Committee Selection There is a club consisting of six distinct men and seven distinct women. How many ways can we select a committee of three men and four women?
 A: There are "$6$-choose-$3$" $=\binom{6}{3}$ ways to select the men, and "$7$-choose-$4$" =$\binom{7}{4}$ ways to select the women. That's because 
$(1)$ We have a group of $6$ men, from which we need to choose $3$ for the committee.
$(2)$ We have $7$ women from which we need to choose $4$ to sit in on the committee. 
Since anyone is either on the committee or not, there is no need to consider order: position or arrangement of those chosen for the committee isn't of concern here, so we don't need to consider permuting the chosen men or women.
So we just *multiply*$\;$ "ways of choosing men" $\times$ "ways of choosing women".
[Recall th Rule of the Product, also known as the multiplication principle.] 
$$\binom 63 \cdot \binom 74 = \dfrac{6!}{3!3!} \cdot \dfrac{7!}{4!3!} = \dfrac{6\cdot 5\cdot 4}{3!}\cdot \dfrac{7\cdot 6\cdot 5}{3!}$$
A: Hint: You can splitt the problem in three parts.
(1) in how many ways can you pick a group of 3 men out of 6.
(2) in how many ways can you pick a group of four women out of 7.
(3) in how many ways can you combine these two groups? ( it's just (1)$\times$(2))
