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Combinatorics with cards

  1. May 13, 2016 #1
    1. The problem statement, all variables and given/known data
    4 cards are dealt from a 52-card deck. How many hands contain 2 distinct pairs?

    2. Relevant equations
    This is an expression I come up with 13C1x4C2x12C1x4C2.

    3. The attempt at a solution
    This is how I approach it.
    From the 13 ranks, I choose 1.
    In this rank, I choose 2 cards from 4.

    From the remaining 12 ranks, I choose 1.
    In this rank, I choose 2 cards from 4.

    Now, I do the following calculations with combinations.

    13C1x4C2x12C1x4C2 = 5616.

    My neighbor told me that it's wrong. He didn't explain why it was wrong. I cannot find any mistakes.
    Please explain what I have done wrong. Thank you very much.
     
  2. jcsd
  3. May 13, 2016 #2
    Your method results in double-counting. Using ##^{13}C_{1} \times ^{12}C_{1}## counts the combination (x,y) twice - once as (x,y) and another time as (y,x), when these hands should considered as the same.
     
  4. May 13, 2016 #3
    Assume that one hand is AA, JJ if the first rank is A and the second rank is J. A can't be counted in the second rank as you stated it.
    Based on your counting, another way would be JJ,AA if the first rank is J. The A rank is still available for the second pair generating a double counting.
     
  5. May 13, 2016 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    You need to choose two distinct ranks from the 13 ranks; then for each chosen rank you need to pick 2 of the 4 cards in that rank.
     
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