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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 $C(26,8)/C(52,26)$ ?

2) 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 2nd problem.

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  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.

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  • $\begingroup$ Thanks! 1) if interest in the probability, then $C(4,2)C(26,8)/C(52,13)$? $\endgroup$ – itproxti Dec 19 '16 at 19:39
  • $\begingroup$ You should divide by $\binom{52}{8}$. $\endgroup$ – angryavian Dec 19 '16 at 20:21
  • $\begingroup$ 2) you're missing $C(4,3)$ 3 suits out of 4. $\endgroup$ – itproxti Dec 20 '16 at 4:32

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