Combinatorics homework question 
The answer is 54912.
This is what I've tried so far: So first you have to pick a rank to occur 3 times so thats 13, now you gotta pick a suit that that rank has, which is now 13 * 4. Now you need to pick that same rank 2 more times. The second time it will be 13 * 3, then the third time it will be 13 * 2. So the expression so far is (13^3)*(4*3*2) but that number is already too high when I try to add the other two cards. What am I doing wrong?
 A: When you pick the first rank for the "3 of a kind", there are 13 options. Now that you have the rank, you must pick 3 cards from the 4 cards of that rank to be the rank which occurs exactly 3 times. There are 4 choose 3 ways of doing that. Now your last two cards must be of a different rank than the first and not of the same rank either. Since there are two ranks left to choose from, there are 12 choose 2 possible ranks for these last two cards. And for each of these two ranks that were selected, there are 4 possible suits for each of the two ranks. So in total we have
$13 \cdot {4 \choose 3} \cdot {12 \choose 2} \cdot 4 \cdot 4 = 54912$.

When you say that "you need to pick the second rank 2 more times", since you have already picked a suit, there are not 13*3 options. After you have picked the first rank (13 options), you then decide on which 3 suits of that rank you will have in the hand. There are 4 options for the first, 3 for the second, and 2 for the third. But then you need to divide by 6=3*2 because the order of the 3 cards in the hand is irrelevant. Thus is just (4 choose 3).
A: The key to this question is the terms 'combination' and 'subset'.  You want to model this problem with combinations for each subset/property and then multiply these together.  
That is: rank $\times$ suit $\times$ other_rank $\times$ other_suits.
One way to think about this is build an example: (4,{diamond,heart,spade},(2,5),{diamond,heart}).
There are 13 ranks so we choose one with $\binom{13}{1}$, choose our suits with $\binom{4}{3}$, chose our other two ranks $\binom{12}{2}$ and then the suits for these two $\binom{4}{1}^2$.
So the answer is $\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2=54912$.
