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Thread: factorial.. with replacement?

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    factorial.. with replacement?




    Here is the problem.. (literally..)
    In your stats class, the teacher has three drawing for prizes (a, b , c). There are 15 students in class

    a. if same person can win more than once, in how many different ways
    can you choose the 3 names from the 15 students.

    I am assuming this is WITH replacement...

    without replacement you would use 15!/ (15-3)! * (3!) or 455
    how do you factor back in the names already picked.. or am I way off
    track?

    b. If the selections above are random, what is the probability that the same
    student wins all 3 prizes..

    is this right (1/15)(1/15)(1/15) = 1/3375 ??

    Thanks for your time.. M

  2. #2
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    a. since you can replace the name after each draw, there are 15*15*15 ways that the names can be drawn

    b. 15 divided by the answer in a. (since any of the 15 students can win all 3)
    --> in other words, if you wrote out all the combinations from a., you would find that there are 15 instances of 1 student winning all 3 prizes

  3. #3
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    Thanks.. yet so obvious now..

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