# Picking exactly $3$ different suits in a $4$ card hand

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!

• This tutorial explains how to typeset mathematics on this site. Jan 14, 2019 at 11:15

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. $$\binom{4}{1}\binom{13}{2}\binom{3}{2}\binom{13}{1}^24!$$