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Math Help - number of pairs

  1. #1
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    number of pairs

    if u are dealt 5 cards what is the possible way of getting a pair which a pair can be a 2 and a 3
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by robkitch View Post
    if u are dealt 5 cards what is the possible way of getting a pair which a pair can be a 2 and a 3

    EDIT:
    Soroban's post gave me greater insight just now

    \frac{ {4 \choose 1} {4 \choose 1}{44 \choose 3} }{{52 \choose 5}} = 0.08153 = 8,15 \ percent \ approx
    Last edited by janvdl; June 18th 2008 at 09:48 AM.
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  3. #3
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    Hello, robkitch!

    Your English is not quite readable . . .


    if u are dealt 5 cards what is the possible way of getting a pair which a pair can be a 2 and a 3
    I'll take a guess at what you meant . . .


    We want five cards: one Deuce, one Trey, and three Others.

    To get one Deuce, there are: {4\choose1} = 4 ways.

    To get one Trey, there are: {4\choose1} = 4 ways.

    To get three Others, there are: {44\choose3} = 13,244 ways.


    Therefore, there are: . 4 \times 4 \times 13,244 \:=\:\boxed{211,904\text{ ways}}

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