Given a standard $52$ card deck, I want to know how many different ways it is possible to pick four cards so that you have exactly $3$ suits in your hand (i.e. there is exactly one suit pair). Additionally, order does matter, so for example: ace of hearts, two of hearts, ace of clubs, ace of spades is distinct from two of hearts, ace of hearts, ace of clubs, ace of spades.
I have two approaches to this problem, but I can't figure out which (or either) is right.
1) $52$ cards to start with. Then $13$ possible cards of one suit, $13$ possible cards of another suit, $12$ possible cards of whichever suit already have. $52 \cdot 13 \cdot 13 \cdot 12 = 105,456$
2) $4!$ combinations. $13$ choose $2$ of one suit, $13$ choose $1$ of another, $13$ choose $1$ of another. $4! \cdot \binom{13}{2}\binom{13}{1}\binom{13}{1} = 316,368$
Thanks!