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Math Help - Choosing 4 cards of different suits and different values from a deck of 52 cards.

  1. #1
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    Choosing 4 cards of different suits and different values from a deck of 52 cards.

    Hello All,

    Please help me solve this problem:
    How many ways are there to choose 4 cards of different suits and different values from a deck of 52 cards?

    Thank you,
    Ashwin Kumar D P, Bangalore, India
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  2. #2
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    Hello, Ashwin!

    I had to "draw a picture" to understand this one . . .


    How many ways are there to choose 4 cards of different suits
    and different values from a deck of 52 cards?
    There are, of course, 52 cards in the deck.

    . . \begin{array}{cccc}<br />
\clubsuit & \heartsuit & \spadesuit & \diamondsuit \\ \hline<br />
A&A&A&A \\ 2&2&2&2 \\ 3&3&3&3 \\ 4&4&4&4 \\ 5&5&5&5 \\ <br />
\vdots & \vdots & \vdots & \vdots \\ K & K & K & K \end{array}


    The first card can be any of the 52 cards.
    . . Suppose it is A\clubsuit
    Then these choices remain for the second card:

    . . \begin{array}{cccc}<br />
\clubsuit & \heartsuit & \spadesuit & \diamondsuit \\ \hline<br />
\boxed{A}&\rlap{X}A&\rlap{X}A&\rlap{X}A \\ \rlap{X}2&2&2&2 \\ \rlap{X}3&3&3&3 \\ \rlap{X}4&4&4&4 \\ \rlap{X}5&5&5&5 \\ <br />
\vdots & \vdots & \vdots & \vdots \\ \rlap{X}K & K & K & K \end{array}

    There are 36 choices for the second card.


    Suppose the second card is 2\heartsuit
    Then these choices remain for the third card:

    . . \begin{array}{cccc}<br />
\clubsuit & \heartsuit & \spadesuit & \diamondsuit \\ \hline<br />
\boxed{A}&\rlap{X}A&\rlap{X}A&\rlap{X}A \\<br />
\rlap{X}2&\boxed{2}&\rlap{X}2&\rlap{X}2 \\<br />
\rlap{X}3&\rlap{X}3&3&3 \\<br />
\rlap{X}4&\rlap{X}4&4&4 \\<br />
\rlap{X}5&\rlap{X}5&5&5 \\ <br />
\vdots & \vdots & \vdots & \vdots \\<br />
\rlap{X}K & \rlap{X}K & K & K \end{array}

    There are 22 choices for the third card.


    Suppose the third card is 3\spadesuit
    Then these choices remain for the fourth card:

    . . \begin{array}{cccc}<br />
\clubsuit & \heartsuit & \spadesuit & \diamondsuit \\ \hline<br />
\boxed{A}&\rlap{X}A&\rlap{X}A&\rlap{X}A \\<br />
\rlap{X}2&\boxed{2}&\rlap{X}2&\rlap{X}2 \\<br />
\rlap{X}3&\rlap{X}3&\boxed{3}&\rlap{X}3 \\<br />
\rlap{X}4&\rlap{X}4&\rlap{X}4&4 \\<br />
\rlap{X}5&\rlap{X}5&\rlap{X}5&5 \\ <br />
\vdots & \vdots & \vdots & \vdots \\<br />
\rlap{X}K & \rlap{X}K & \rlap{X}K & K \end{array}

    There are 10 choices for the fourth card.


    Disgarding the order in which the four cards are drawn,

    . . there are: . \frac{52\cdot36\cdot22\cdot10}{4!} \;=\;\boxed{17,160} such 4-card hands.

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  3. #3
    Junior Member
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    I understand where the 52, 36, 22, and 10 came from, but why divide by 4!? could anyone describe it in a way that makes sense?
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