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Math Help - If you are dealing with a standard deck of cards-how many...data management

  1. #1
    Junior Member risteek's Avatar
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    If you are dealing with a standard deck of cards-how many...data management

    I don't know where this belongs but this seems close at least.

    My partner and I have been going at this for over an hour. We are using nCr ish forumulas but this one doesn't seem to work...

    If you are dealing from a standard deck of cards,
    a)how many different 4-card hands could have at least one card from each suite
    b)how many different 5-card hands could have at least one spade?
    c) how many different 5-card hands could have at least 2 face cards(jacks, queens or kings)?

    I am sorry if this is in the wrong board if if the title sucks. But thanks!
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  2. #2
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    Hello, risteek!

    You are dealing from a standard deck of cards,
    a) How many 4-card hands have at least one card from each suit?
    Strange wording . . . the "at least" is unnecessary.

    We must have one card from each suit.

    There are: 13 choices for the Club, 13 choices for the Heart,
    . . 13 choices for the Spade, and 13 choices for the Diamond.

    There are: . 13^4 \:=\:28,\!561 possible hands.




    b) How many 5-card hands have at least one spade?
    There are 13 Spades and 39 Others.

    There are: . _{52}C_5 \:=\:2,\!598,\!960 possible 5-card hands.

    How many have no Spades?
    Then the five cards must be chosen from the 39 Other cards.
    There are: . _{39}C_5 \:=\:575,\!757 ways.

    There are: . 2,\!598,\!960 - 575,\!757 \:=\:2,\!023,\!203 hands with at least one Spade.




    c) How many 5-card hands have at least 2 face cards?
    There are 12 Face cards and 40 Others.

    The opposite of "at least 2 face cards" is "0 or 1 face card."

    0 Face cards
    The 5 cards are chosen from the 40 Others.
    There are: . _{40}C_5 \:=\:658,008 ways to get 0 Face cards.

    1 Face card
    There are: . _{12}C_1 = 12 ways to get one Face card.
    There are: . _{40}C_4 = 91,\!390 to get 4 Others.
    . . There are: . 12\cdot91,\!390 \:=\:1,096,680 ways to get one Face card.

    Hence, there are: . 658,\!008 + 1,\!096,\!680 \:=\:1,\!754,\!688 ways to get 0 or 1 Face card.


    Therefore, there are: . 2,\!598,\!960 - 1,\!754,\!688 \:=\:844,\!272 hands with at least 2 Face cards.

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  3. #3
    Junior Member risteek's Avatar
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    Thank you so much! You are amazing!!
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