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How many bridge hands have exactly two 5-cards suits and a void so that the remaining suit has a run of 3 cards?

Should the answer be $C(13,5)C(13,5)C(4,2)C(13,3)$.I first choose $5$ cards for each of the $5$-card suits,then choosing $2$ suits from $4$ suits for the two $5$-card suits.Since there's a void in the bridge hands,the remaining cards are of the same suit.I just need to choose the remaining $3$ cards from $52-13-13-13=13$ cards.

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    $\begingroup$ also you need to pick which of the two remaining suits has the void, so multiply by C(2,1)=2 $\endgroup$
    – WW1
    Commented May 14, 2015 at 15:58
  • $\begingroup$ But like the question here: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)? The answer is :C(12,4)C(4,1)C(13,4)C(3,1)C(13,4) I don't need to pick up which of the two remaining suits has the void.Why's that? $\endgroup$
    – JimfyWinsy
    Commented May 16, 2015 at 5:49
  • $\begingroup$ I think you do need to pick up which of the remaining suits in that case also $\endgroup$
    – WW1
    Commented May 16, 2015 at 17:27
  • $\begingroup$ I take it that a run of $3$ cards here just means any three cards, like you have taken in your attempt $\endgroup$
    – true blue anil
    Commented Jun 23, 2022 at 18:08

1 Answer 1

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To avoid confusion evident in comments, I follow this practice

  • write the # of cards distributed to the hands in descending order
  • permute the hands
  • distribute cards to the hands

For your problem the pattern is $5-5-3-0$ with $\frac{4!}{2!1!1!}$ permutations,
so the answer is $\dfrac{4!}{2!1!1!}C(13,5)C(13,5)C(13,3)$

And for the problem in your comments,
the pattern is $5-4-4-0$, with $\frac{4!}{1!2!1!}= 12$ permutations,
and with the ace condition attached,
the distribution is $C(12,4)C(13,4)C(13,4)$

so clearly $C(4,1)C(3,1)=12$ shouldn't be multiplied by $C(2,1)$

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