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Math Help - Counting

  1. #1
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    Counting

    Ace high hands

    \dbinom{4}{1}*\dbinom{12}{4}*\left[\dbinom{4}{1}\right]^4=506880

    Is this correct because someone else said the answer is 502,860.
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  2. #2
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    Hello, dwsmith!

    Your answer is off . . . and Someone Else's answer is off, too.


    Ace high hands

    \displaystyle \binom{4}{1}\cdot\binom{12}{4}\cdot\left[\binom{4}{1}\right]^4\:=\:506,\!880

    Is this correct because someone else said the answer is 502,860.

    Your answer is correct . . . up to a point.

    You said there are 4 choices for the Ace
    . . then we select 4 of the other 12 values
    . . and there are 4^4 ways to select those values.

    You found the number of hands which contain an Ace and no pairs. . Good!


    However, this includes the hands that are Straights or Flushes.


    There are Ace-high straights: . \{10,J,Q,K,A\}
    . . There are: . {4\choose1}^5 \:=\:1024 of them.

    There are Ace-low straight: . \{A,2,3,4,5\}
    . . There are: . {4\choose1}^5 \:=\:1024 of them.

    Hence, there are: . 1024 + 1024 \:=\:2048 Straights.


    For the Flushes, there are: {4\choose1} choices for the Ace.
    And {12\choose4} choices for the other 4 cards of that suit.

    Hence, there are: . {4\choose1}\cdot{12\choose4} \:=\:1980 Flushes.


    It seems that there are: . 1024 + 1980 \:=\:4028 Straights and Flushes.

    And this is what Some Else subtracted.


    But he/she forgot that we've overcounted.
    There are some hands which are Straights and Flushes.

    There are 4 hands that are Ace-high straight flushes.
    (They're called Royal Flushes, remember?)

    So, the number we subtract is: . 1980 + 2048 - 4 \:=\:4024


    Therefore, there are: . 506,\!880 - 4,\!024 \;=\;502,\!856 Ace-high hands.

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  3. #3
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    What are your asking about?
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