combination problem. A small company employs $3$ men and $5$ women. If a team of $4$ employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly $2$ women?
So, first, how many possible combinations are there. Well there are $8$ people and a group of $4$ people are going to be selected. We want to know how many of these teams can be selected with $2$ women. It seems like order does not matter and so its a combination problem. MWMW is the same as MMWW here. 
Anyway, how many total combinations are there are 8 ppl in 4 teams? Well its:
$_8C_4 = \frac{8!}{4!4!} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}=
\frac{2 \cdot 7 \cdot 2 \cdot 5}{1 \cdot 1 \cdot 2 \cdot 1}= \frac{140}{2} = 70$.
So there are $70$ total groups. So that goes in the denominator for the answer. What's our numerator?
How do I proceed from here?

There are $8$ employees. $4$ are chosen from the $8$ to organize the retreat: there are ${8 \choose 4} = 70 $ ways to do this. How many of these ways have just $2$ women? Well, remember that we have $5$ women. There are ${5 \choose 2} = 10$ ways to pick them. Then we have $3$ men. If in a team of $4$ we have $2$ women, then we must have $2$ men. There are ${3 \choose 2} = 3$ ways to pick them. By the product rule, we have $10 \cdot 3 = 30$ different ways of picking exactly $2$ men and $2$ women. So there are $30$ teams with exactly $2$ women in them. So the probability of picking such a team is $\frac{30}{70}=\frac37$.
