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.

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 
$$\frac{4\cdot\binom{3}{2}\cdot\binom{13}2\cdot 13\cdot 13}{\binom{52}4}=\frac{26\cdot\binom{13}{2}^2}{\binom{52}4}.$$
