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. $4! \cdot \binom{13}{2}\binom{13}{1}\binom{13}{1} = 316,368$


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1 Answer 1


There are four ways to select the suit from which two cards are drawn, $\binom{13}{2}$ ways to select two of the thirteen cards of that suit, $\binom{3}{2}$ ways to select two of the remaining three suits, thirteen ways to select a card from each of those suits, and $4!$ ways to arrange the selected cards. $$\binom{4}{1}\binom{13}{2}\binom{3}{2}\binom{13}{1}^24!$$

Your first approach is wrong since the second card of the same suit may be picked with the second, third, or fourth card. Also, you have not selected from which suits the cards will be drawn.

Your second approach is wrong since you have not selected the suits from which the cards will be drawn.


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