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How many bridge hands have a 5-card suit that must contain the ace of that suit, a 4-card suit, and a void (no cards of a suit)? I only know the first two are: C(12,4)C(13,4), but I really don't know what the "void" means and how to deal with it in the question.

Thanks!

ReliableMathBoy
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JimfyWinsy
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1 Answers1

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The distribution is of the type $5$-$4$-$4$-$0$. (We are void in a suit, say spades, if we have $0$ spades.)

The suit we have $5$ of can be chosen in $\binom{4}{1}$ ways. For each such choice, as you saw, the actual cards can be chosen in $\binom{12}{4}$ ways, since one of the cards must be an Ace. For each such choice, the suit we are void in can be chosen in $\binom{3}{1}$ ways. And then the cards in the two suits we have $4$ of can be chosen in $\binom{13}{4}^2$ ways. Multiply.

André Nicolas
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  • Why we don't need to choose 1 suit for each of the 4-card suit?Just as what you did for the 5-card suit and void. – JimfyWinsy May 03 '15 at 04:08
  • Once we have chosen which suit has $5$ cards, and which suit is missing, the two suits we have $4$ cards in are determined. – André Nicolas May 03 '15 at 04:36
  • But the 4 cards must be in one same suit,not two. – JimfyWinsy May 15 '15 at 13:47
  • The problem says we have $5$ cards in one suit, $4$ in another, and a void ($0$ cards) in another. That only accounts for $9$ cards. But a bridge hand has $13$ cards. Thus there must be $4$ cards in the fourth suit. – André Nicolas May 15 '15 at 13:51
  • But like the problem here:how many bridge hands have exactly two 5-cards suits and a void so that the remaining suit has a run of 3 cards? The answer is :C(13,5)C(13,5)C(4,2)C(2,1)C(13,3). They are exactly the same type of question,but we need to multiply C(2,1).Why's that? – JimfyWinsy May 16 '15 at 18:22
  • Not exactly the same. For the problem you just mentioned, we have a 5-5-3-0. The suits we have $5$ of can be chosen in $\binom{4}{2}$ ways. For each the cards can be chosen in $\binom{13}{5}$ ways. Now the suit we have $3$ of can be chosen in $\binom{2}{1}$ ways. Now we need to interpret run of $3$ cards. I would think it means something like 10, Jack, Queen. Then it depends on whether Ace can count as both high and low, or just high. So either multiply by $11$ or $10$. Your $\binom{13}{3}$ has us choosing arbitrary cards. – André Nicolas May 16 '15 at 18:41