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Math Help - Counting Principle 52-card deck

  1. #1
    DBA
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    Counting Principle 52-card deck

    How many ways are there to pick 2 different cards from a standard 52-card deck such that:

    The fist card is a spade and the second card is not a Queen

    The answer is (1*48) + (12*47)

    I do not know how to get there again.

    I have 13 spade cards in a 52-card deck
    Ace, 2,3,4,5,6,7,8,9,10, Jack, Queen, King

    so I thought there are 13 possibilities for the first card
    For the second card we cannt have Queens (4) and one card is already gone,
    so that would be 52-4 - 1 = 47

    I see the problem that maybe for the first card a spade Queen was used, but I do not know how to set the problem up with this information.

    Thanks for any hint.
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  2. #2
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    Hello, DBA!

    How many ways are there to pick 2 different cards
    from a standard 52-card deck such that:

    The first card is a Spade and the second card is not a Queen.

    The answer is: .  (1\cdot48) + (12\cdot47)

    We must consider two cases:

    . . (1) the first card is the Q\spadesuit.

    . . (2) the first card is not the Q\spadesuit.


    (1) The first card is the Q\spadesuit.
    . . . There is one way to accomplish that.

    . . . Among the other 51 cards, there are 3 Queens and 48 Others.
    . . . And we want one of the Others: 48 ways.

    . . . Hence: . n(\text{1st}\,Q\spadesuit \:\wedge \sim\!Q) \;=\;1\cdot48


    (2) The first card is a \spadesuit, but not the Q\spadesuit.
    . . . There are 12 ways to accomplish that.

    . . . Among the other 51 cards, there are 4 Queens and 47 Others.
    . . . And we want one of the Others: 47 ways.

    . . . Hence: . n(\text{1st} \sim\! Q\spadesuit\;\wedge\,\sim\!Q) \;=\;12\cdot47


    Therefore: . n(\text{1st}\spadesuit\;\wedge\:\sim\!Q) \;=\;1\cdot48 + 12\cdot47 \;=\;612\text{ ways.}

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