# How many eight-card hands can be chosen from exactly 2 suits/13-card bridge hands contain six cards one suit and four and three cards of another suits

1. How many eight-card hands can be chosen from exactly $$2$$ suits of an ordinary $$52$$-card deck? (there are $$4$$ suits clubs, diamonds, hearts and spades

I think since there are $$26$$ cards in $$2$$ suits and eight cards from those $$26$$ (order does not matter), thus $$\displaystyle\frac{\binom{26}{8}}{\binom{52}{26}}$$ ?

1. How many $$13$$-card bridge hands can be chosen from an ordinary $$52$$-card deck that contain six cards of one suit and four and three cards of another two suits? (there are $$4$$ suits clubs, diamonds, hearts and spades

I do not understand $$2$$nd problem.

1. I don't know why you are dividing by $\binom{52}{26}$. The number $\binom{26}{8}$ already counts the number of 8-card hands you could get from a pre-specified $26$ cards. You may need to multiply by $\binom{4}{2}$ to choose which two suits to use, though.
2. There are $4 \cdot 3 \cdot 2$ ways to choose the 6-card suit, the 4-card suit, and the 3-card suit to make up your hand. Then you multiply by $\binom{13}{6}$, $\binom{13}{4}$, and $\binom{13}{3}$ to choose the number of cards of each suit.
• Thanks! 1) if interest in the probability, then $C(4,2)C(26,8)/C(52,13)$? Commented Dec 19, 2016 at 19:39
• You should divide by $\binom{52}{8}$. Commented Dec 19, 2016 at 20:21
• 2) you're missing $C(4,3)$ 3 suits out of 4. Commented Dec 20, 2016 at 4:32