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I have a deck of 12 cards - one Jack, Queen and King of each suit. There are 5 cards in one hand. How many hands are there in which a Jack, Queen and King all show up and all 4 suits show up? My thinking behind this:

Number of hands = (Number of ways to choose the 3 suits that we get the Jack, Queen and King from) $\times (3!) \times$ (Number of ways to choose any card from the 4th suit) $\times$ (Number of ways of getting any 5th card) = $\binom{4}{3}\times3!\times\binom{3}{1}\times\binom{8}{1} = 576$.

I'm not sure if this is right as the total seems too large. Please confirm/correct where I am going wrong, thanks.

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  • $\begingroup$ You need to start accepting answers to questions you have already asked. $\endgroup$ – Derek Luna Apr 18 at 23:10
  • $\begingroup$ Apologies, just done it. $\endgroup$ – Azamat Bagatov Apr 18 at 23:20
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First let's count the hands that have all four suits, regardless of the ranks. You have to have two cards of one suit and one of each of the others. There are three ways to choose the card(s) in each suit, so there are $4\cdot 3^4=324$ hands.

Now let's count how many of those do not have all three ranks. There are $12$ ways to choose the suit with two cards and the two cards in that suit. Then to miss a rank there are $2$ choices in each other suit, so there are $12\cdot 2^3$ hands with all four suits and not all three ranks.

That leaves $228$ hands with all four suits and all three ranks.

Your approach to counting double counts many hands. A hand with the Jacks of hearts and spades will be counted at least twice, once for each Jack being the first card and the other being the last. It may also get counted with neither being the first card, but these two being fourth and fifth.

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