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Difficult Math Question #67 - Combinations

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Question #67 - Combinations

Post Wed Nov 29, 2006 2:49 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    A shipment of 10 TV sets includes 3 that are defective. In how many ways can a hotel purchase 4 of these sets and receive at least two of the defective sets?

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    lalitaroral Junior | Next Rank: 30 Posts Default Avatar
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    Post Wed Nov 29, 2006 10:22 pm
    D:Defect, C:Correct

    Possible combinations to select 4 TV(condition atleast 2 should be defective)

    1. 3D 1C
    2. 2D 2C

    Case 1: 7C1*3D3
    Case 2: 7C2*3D2

    Total Possible ways: 7C1*3D3 + 7C2*3D2 = 84 Ways

    kulksnikhil Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Dec 01, 2006 3:55 am
    800guy wrote:
    A shipment of 10 TV sets includes 3 that are defective. In how many ways can a hotel purchase 4 of these sets and receive at least two of the defective sets?
    Possible ways

    1> All 3 defective and 1 correct
    2> 2 defective and 2 correct


    1> 1 * 7 = 7
    2> 3C2 * 7C2 = 6 * 21 = 126

    total= 126 + 7 = 133

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Fri Dec 01, 2006 5:26 pm
    OA:

    There are 10 TV sets; we have to choose 4 at a time. So we can do that by 10C4 ways. We have 7 good TV's and 3 defective.

    Now we have to choose 4 TV sets with at least 2 defective. We can do that by

    2 defective 2 good
    3 defective 1 good

    That stands to 3C2*7C2 + 3C3*7C1 (shows the count)

    If they had asked probability for the same question then

    3C2*7C2 + 3C3*7C1 / 10C4.

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