odds of picking exactly 2 women and 2 men out of 12 men and 12 womem I understand the answer to be 12 choose 2 * 12 choose 2 over 24 choose 4. 
I don't really understand why, or what principle I can extract from the problem. I can understand that we are putting the total possible outcomes in the denominator, but not how the numerator represents the exact 2 by 2 requirement the problem states
 A: There are ${12}\choose{2}$ ways to pick $2$ women out of $12,$ without regards to order (i.e. Mary & Lucy $=$ Lucy & Mary). For each of these ways, there are ${12}\choose{2}$ ways to choose $2$ men. So you must multiply to get the number of possibilities: ${12}\choose{2}$ $\cdot$ ${12}\choose{2}$. Assuming equal likelihood of any choice, divide this by the total number of outcomes ${24}\choose{4}$ to get the probability of any particular group of $4.$
A: The fact that you understand the denominator is a good start.
There isn't much "theory" to get from this problem, but you should simply look at each event in isolation (men, then women) and combine them.
The number of ways of choosing 2 men out of 12 men is $\binom{12}{2}$ or as you said it in words "12 choose 2". The same goes for how many ways to choose exactly 2 women out of 12.
Now think of it this way. In math we usually associate "and" probabilities by multiplying them together, and you can sort of think in the same way here. If you think of Event A as the choosing 2 men and event B as choosing 2 women, for both A and B to occur you want to multiply the number of ways that each can happen.
Said another way, if there are 100 ways to choose men and 100 ways to choose women, then there are 100*100=10,000 ways to choose them both. For each "way" of choosing men, I still have 100 ways of choosing the women to pair with them. That's where the multiplication comes from.
The denominator is just the total number of ways of choosing 4 people out of 24
