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Homework Help: Dealt foru cards, how many hands consit of 2 pairs? i have the answer

  1. Nov 14, 2006 #1
    Hello everyone. I was loooking over some notes and I noticed this problem i had a question mark next too.

    Supose that you are dealt four cards instead of 5. How many 4 card hands consist of two pairs (two cards of one denomination and two cards of a second denomoniation)?

    The answer is 2,808.

    It was found by:
    (13 choose 2) * (4 choose 2)^2

    But i'm confused because if you are choosing 2 cards of one denomination, thats like choosing two 5's. There is only one 5 in every 13 cards or suites correct?

    so if you have
    (13 choose 2) that means your choosing 2 cards out of 1 suit weither it be hearts, diamonds, clubs or spades. So how can you be choosing 2 of the same numbers from a set of 13?

    Or are they saying, they are chooosing 2 cards of 1 suit, such as a 4 and a 10, and later they are choosing 2 cards to match that suit, such as a 10H and a 10C, that would be one (4 choose 2) and the other would be 4H, and a 4D that would be the other (4 choose 2)?

    I think i got it by writing this question but just making sure, thanks!
  2. jcsd
  3. Nov 14, 2006 #2


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    Let nCr mean "n choose r".

    It's 13C2 because we need to choose 2 numbers that will be the two doubles, and there are only 13 numbers to choose from. Hence, we choose 2 from the 13 numbers.

    Now, for the first number we have chosen, we must choose 2 suits out of four. That's your standard deck, so we have 4C2. Similarly for the second card we have chosen for the other double, so we have 4C2 again.

    Hence, (13C2)(4C2)(4C2) = (13C2)(4C2)^2.
  4. Nov 14, 2006 #3
    Ahh I get it now, thanks again!
  5. Nov 14, 2006 #4


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    I'm glad you got it.

    Now, I'm having a hard time with my Probability assignment myself. :frown:
  6. Nov 14, 2006 #5
    Post it up! hah i doubt I could help though, at least i'll be ready for my stat classes coming up from all this probability in descrete math, wee.
  7. Nov 14, 2006 #6


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    Yeah, I should be ready for Stats next term though.

    Not sure how far your Probability goes, but it does go into complicated territory.

    I will be posting up questions soon. :smile:
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