If I draw 4 cards from a deck of 52 cards, what is the probability that the cards are 2 of one suit and 2 of another? 4 cards are drawn from a stack of 52, what is the probability that 2 cards are of one suit and the other 2 cards are of a different suit? 
I.e. 2 Diamonds and 2 Clubs or 2 Spades and 2 Diamonds etc..
I think it might be: 
$$2\left(\frac{13C2 \cdot 13C2}{52C4}\right)$$
However, I'm not entirely sure that multiplying it by 2 to account for all possible suits is the correct approach.
 A: In my opinion "2 cards are of one suit and the other 2 cards are of a different suit" is a bit ambiguous. Let us consider the following two possible interpretations.
2 different suits: two cards of one  suit and two of another suit 
The total number of hands of $4$ cards is $\binom{52}{4}.$
We have $\binom{4}{2}$ ways to choose the suits. Then we have $\binom{13}{2}$ ways for the values of the suit with two cards and $\binom{13}{2}$ for the other two. 
So the probability should be 
$$\frac{\binom{4}{2}\cdot\binom{13}{2}^2}{\binom{52}4}=\frac{6\cdot\binom{13}{2}^2}{\binom{52}4}.$$
3 different suites: two cards of one  suit and two of the other suits
The total number of hands of $4$ cards is $\binom{52}{4}.$
We have $4$ ways to choose the suit with two cards and $\binom{3}{2}$ to choose the other two suits. Then we have $\binom{13}{2}$ ways for the values of the suit with two cards and $13\cdot 13$ for the other two. 
So the probability should be 
$$\frac{4\cdot\binom{3}{2}\cdot\binom{13}2\cdot 13\cdot 13}{\binom{52}4}=\frac{26\cdot\binom{13}{2}^2}{\binom{52}4}.$$
A: I think, much as it may seem a little laborious, the most definite way you can do this is by drawing a tree diagram. You can draw out all the possible hands and calculate the probabilities that way. It is less sophisticated than the methods above using $^nC_r$, but also allows for less error. I'll attach an image of the diagram I've drawn up below and then talk through it.

Let the first card drawn have suit X, and the first card drawn not in suit X have suit Y. (Any subsequent suit drawn will be called Z).
Whatever the first card drawn is, that will be suit X. The next draw (the second column) will be one of two possibilities: either the card will be in the same suit as the first (going up from the first box, with a probability of $\frac{12}{51}$), or it will be different (going down, with a probability of $\frac{39}{51}$).
The next draws get a little more complicated. Let's take the scenario that we have two cards of the same suit first. There are two possibilities to follow: either the next card is also of the same suit (which we don't want), or it is of a different suit (which has a probability of $\frac{39}{50}$ (in the diagram I have only shown probabilities that lead to a successful outcome). From this, there is only one successful possibility - card 4 has the same suit as card 3 (a probability of $\frac{12}{49}$).
Let's now take the other half of the diagram, where card 1 and card 2 were different suits (the lower branch). Draw three has three possible outcomes:


*

*Card 3 is the same suit as card 1 (a probability of $\frac{12}{50}$).

*Card 3 is the same suit as card 2 (a probability of $\frac{12}{50}$).

*Card 3 is of a different suit (which won't lead to a successful situation).


For possibility 1, we currently have {X, Y, X}, and so to fulfil the criteria, we want another Y (the probability of drawing which is $\frac{12}{49}$). For possibility 2, we currently have {X, X, Y}, and so to fulfil the criteria, we want another X (the probability of this is also $\frac{12}{49}). We have now got all the possible routes to the answer.
Therefore, the probability (by the addition principle) will be:
$P(2\ \textrm{suit}\ 1, 2\ \textrm{suit}\ 2) = \Big(\frac{12}{51} \times \frac{39}{50} \times \frac{12}{49} \Big) + \Big(\frac{39}{51} \times \frac{12}{50} \times \frac{12}{49} \Big) + \Big(\frac{39}{51} \times \frac{12}{50} \times \frac{12}{49} \Big)$
From writing out the calculation, we now can see that the probability can be simplified to:
$P(2\ \textrm{suit}\ 1, 2\ \textrm{suit}\ 2) = 3 \times \Big(\frac{12^{2} \times 39}{51 \times 50 \times 49} \Big) = \frac{6 \times 12 \times 39}{17 \times 25 \times 49} = \frac{2808}{28025}$
which is equal to DreamConspiracy's answer of $\displaystyle 6 \Bigg( \frac{{13 \choose 2}^2}{{52 \choose 4}} \Bigg)$.
Hope this helps!
A: The right part of your expression is already correct. If we assume that you have already picked two suits, then there are indeed $\frac{\binom{13}2^2}{\binom{52}4}$ ways of picking these cards. So how do we pick two suits? Well, it's the same approach. You have a total of four suits and need to pick two. That's just $\binom 42=6$, giving you a final answer of $6\left(\frac{\binom{13}2^2}{\binom{52}4}\right)$.
