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Math Help - Ball and Urn Model

  1. #1
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    Ball and Urn Model

    I need to find the probability of calculating placing of r balls into n urns, such that there can be no empty urns.

    I find the number of ways to do it is {n-1}\choose{r-1}.
    But this is pretty useless. It does not give me the actual probability.

    For example, for n=2, r=3.
    We can list down all the cases:
    ABC|x
    x|ABC
    AB|C
    AC|B
    CB|A
    A|BC
    B|AC
    C|AB

    The probability is 6/8.
    The number of ways is 6.

    Using the {n-1}\choose{r-1} formula, I get:
    1\choose2, which is pretty much meaningless.

    Can I use any mathematical models to get the number 6, instead of having to list all down?

    Thanks!
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  2. #2
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    If you have read my response to an earlier question, then you know that you are using a model in which objects-distinguishable, cells-distinguishable and cells may not be empty. Putting R distinguishable balls into N distinguishable urns without any urn being empty of course requires R \ge N. The number of ways to do that is simply the number of surjections (onto functions) from a set of R elements to a set of N elements. I will give you the counting formula for that: <br />
Surj(R,N) = \sum\limits_{k = 0}^N {\left( { - 1} \right)^k  { N \choose k } \left( {N - k} \right)^R }.
    To understand this idea, you may want to study Stirling numbers of the second kind.
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  3. #3
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    Thanks! That was very helpful indeed, throwing me a lead on Stirling's Number.
    The amount of knowledge to master for probability is really vast.
    Last edited by chopet; November 17th 2007 at 07:01 AM.
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