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Math Help - combinations and permutations problem

  1. #1
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    combinations and permutations problem

    The question is: How many cards must be drawn (without replacement) from a standard deck of 52 to guarantee the following?

    a) Two of the cards will be of the same suit
    b) Three cards of the same suit

    Since both questions are set up the same way I would appreciate help with either of those.

    I really have trouble getting to a formula from a word problem. And, the only numbers I think I have to work with are 52, 13, and 2. I know that the answer is 5 as I've already read that you can draw 4 cards and the 5th card would be the 2nd card of one of the previously drawn suits. (because there are not 5 suits)

    I guess my question is: are they looking for a formula? How would one set this up? Thank you for your consideration.
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  2. #2
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    Quote Originally Posted by robitussin217 View Post
    The question is: How many cards must be drawn (without replacement) from a standard deck of 52 to guarantee the following?

    a) Two of the cards will be of the same suit
    b) Three cards of the same suit

    Since both questions are set up the same way I would appreciate help with either of those.

    I really have trouble getting to a formula from a word problem. And, the only numbers I think I have to work with are 52, 13, and 2. I know that the answer is 5 as I've already read that you can draw 4 cards and the 5th card would be the 2nd card of one of the previously drawn suits. (because there are not 5 suits)

    I guess my question is: are they looking for a formula? How would one set this up? Thank you for your consideration.
    There really isn't need to set up a formula here -- just use logic: Since there are 4 suits, you can have 4 cards all of different suits, however in any 5-card combination there will be two cards of the same suit. If you insist on a formula, then you will need
    \frac{\text{Number of cards in deck}\cdot [\text{Number of cards of same suit-1}]}{\text{Number of cards per suit}} + 1

    In the first case, this will yield \frac{52 \cdot (2-1)}{13}+1 = 4 + 1 = 5

    In the second, we'll get \frac{52 \cdot (3-1)}{13}+1 = 9, since we can have 8 cards and they will be 2 of the same suit, however any 9 card combination will have 3 cards from some suit.
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  3. #3
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    Hello, robitussin217!

    Another approach . . .


    How many cards must be drawn (without replacement) from a standard deck of 52
    to guarantee the following?

    a) Two of the cards will be of the same suit
    Consider the "worst case scenario" . . .

    If you draw four cards, they could be one of each suit.

    . . Example: . 7\spadesuit,\;3\heartsuit,\;A\clubsuit,\;K\diamond  suit

    The fifth card must give us two cards of the same suit.




    b) Three cards of the same suit

    If you draw eight cards, they could be two of each suit.

    . . Example: . 8\spadesuit,3\spadesuit,\;5\heartsuit,Q\heartsuit,  \;4\clubsuit,K\clubsuit,\;9\diamondsuit,A\diamonds  uit

    The ninth card must gives us three cards of the same suit.

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