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How many combinations that have at least 2 queens are possible?

  1. Dec 13, 2005 #1
    From a deck of 52 cards, the 12 face cards are removed. From these face cards, 4 are chosen. How many combinations that have at least 2 queens are possible?

    The answer is 201... I can't get here. This is what I did:
    case1) 4 queens can be chosen C(4,2) ways * 10 cards can be chosen C(10,2) ways
    case2) 4 queens can be chosen C(4,3) ways * 9 cards can be chosen C(9,1) ways
    case3) 4 queens can be chosen C(4,4) ways * 8 cards can be chosen C(8,0) ways

    case 1 = 270
    case 2 = 36
    case 3 = 1
    270+36+1 does not equal 201

    what am I doing wrong? TiA
     
    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Dec 14, 2005 #2

    JasonRox

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    Here is your mistake.

    I'll show for Case 1, then you can solve the rest.

    4 Queens can be chosen, which is 4 C 2 ways. The problem is here. You said that 10 cards can be chosen, but that means a Queen can be chosen. That's wrong you can't choose a Queen again. So, it is 2 cards can be chosen from 8, which is 8 C 2.

    See where this is going?

    Now, solve it. :biggrin:
     
  4. Dec 14, 2005 #3

    Tide

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    HINT: If exactly two queens are drawn, how many non-queens are available to complete the foursome?
     
  5. Dec 17, 2005 #4
    If you do this, It's better to do it the reverse way. Find the universal amount of cards and subtract the complement, which is Tide's method: how many non-queens are available to complete the foursome. Most questions like these, if you take a backwards approach to it, it's alot easier.

    Cases can get a bit confusing.

    So:

    n(u) = 12C4

    Let A be the # of sequences that have only 1 queen

    n(a) = 9C4

    Let B be the # of sequences that have no queen

    n(b) = 8C4

    Total number of combinations with at least 2 queens = n(u) - n(a) - n(b)
    = 12C4 - 9C4 - 8C4
     
    Last edited: Dec 17, 2005
  6. Dec 18, 2005 #5
    Thanks ... I feel pretty stupid making a mistake like that.

    ...forevergone, your cases don't work out... I get an answer of 299 =p
     
  7. Dec 20, 2005 #6
    It's probably a mathematical error, but the approach should be like that. If you use the complement and subtract it from the universe, you should get the right answer.
     
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