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Math Help - numbers chosen without.replacement

  1. #1
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    numbers chosen without.replacement

    r numbers are chosen at random 0<r<10 without replacement from the numbers 0,1,2,...,9. What is the probability that no two numbers are same ?
    Answer is 10!/((10^r)*(10-r)!)
    please explain reasoning
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  2. #2
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    Re: numbers chosen without.replacement

    Quote Originally Posted by ice_syncer View Post
    r numbers are chosen at random 0<r<10 without replacement from the numbers 0,1,2,...,9. What is the probability that no two numbers are same ?
    Answer is 10!/((10^r)*(10-r)!)
    If you have posted the question correctly, then the probability is one.
    Without replacement no two numbers can be the same.
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  3. #3
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    Re: numbers chosen without.replacement

    Hello, ice_syncer!

    Please check the wording of the problem.


    r numbers are chosen at random (0<r<10) without replacement ??
    from the numbers 0, 1, 2, ..., 9.
    What is the probability that no two numbers are same ?

    Answer is: 10!/[(10^r)*(10-r)!]

    The given answer is for with replacement.


    Examples:

    P(3) \:=\:\frac{10}{10}\cdot\frac{9}{10}\cdot\frac{8}{1  0} \;=\;\frac{10!}{10^3\,7!}

    P(4) \;=\;\frac{10}{10}\cdot\frac{9}{10}\cdot\frac{8}{1  0}\cdot\frac{7}{10} \;=\;\frac{10!}{10^4\,6!}
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  4. #4
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    Re: numbers chosen without.replacement

    That makes sense. It is possible that there is a printing error in my book. Thanks.
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