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Math Help - Configuration of particles

  1. #1
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    Configuration of particles

    Suppose that n distinguishable particles are placed randomly in N boxes (states). A particular configuration of this system is such that there are ns particles in state s, where 1<=s<=N. If the ordering of particles in any particular state doesn't matter, find the number of ways of realising a particular configuration.

    It's not at all obvious to me what I should do to tackle this question, any help would be hugely appreciated.
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  2. #2
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    Quote Originally Posted by free_to_fly View Post
    Suppose that n distinguishable particles are placed randomly in N boxes (states). A particular configuration of this system is such that there are ns particles in state s, where 1<=s<=N. If the ordering of particles in any particular state doesn't matter, find the number of ways of realising a particular configuration.

    It's not at all obvious to me what I should do to tackle this question, any help would be hugely appreciated.
    Hi Free To Fly,

    If I understand your question correctly, the number of ways is the multinomial coefficient,

    \binom{n}{n_1 \; n_2 \; \dots \; n_N} = \frac{n!}{n_1! \, n_2!  \, \cdots \, n_N!}.
    Last edited by awkward; November 28th 2008 at 03:15 PM. Reason: corrected typo
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  3. #3
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    Thanks for the answer, but could you please explain where you got it from? I can't really see that. Thanks.
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  4. #4
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    The multinomial coefficient

    \binom{n}{n_1 \; n_2 \; \dots \; n_N}

    counts the number of ways to put n distinct objects in N boxes, with n_1 in the first box, n_2 in the second box, etc. See Multinomial theorem - Wikipedia, the free encyclopedia.

    For a derivation, consider all n! permutations of the n objects and consider the first n_1 to be in the first box, the next n_2 to be in the second box, etc. The order of the objects in the boxes is considered irrelevant, however, so divide by n_1! to compensate for over-counting in box 1, divide by n_2! to compensate for over-counting in box 2, ...

    It's just like the binomial coefficient, only generalized to N choices instead of 2.
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