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Counting problem.

  1. Feb 27, 2004 #1
    I think i got this answer.

    How many ways are there to distribute five distinguishable objects into three indistinguishable boxes?

    Wouldn't the answer just be [tex]C(5,3)[/tex] because the boxes are indistinguishable? Or do i treat this question the same as if the boxes were distinguishable?
  2. jcsd
  3. Feb 27, 2004 #2


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    The binomial coefficient [tex]C_3^5[/tex] is right. You could do it as if the boxes were distinguishable and then divide by the number of sequences of them: 3! = 6. and that's what you would get.
  4. Feb 27, 2004 #3


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    Multiple objects can fit in a box, can't they?
  5. Feb 27, 2004 #4
    Unless they are very, very big.
  6. Feb 27, 2004 #5
    Yes, multiple objects can fit into a box.

    I think my answer is correct, isn't it?
  7. Feb 27, 2004 #6

    matt grime

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    I don't think it is. I think it's quite a way out. The answer is the same as the number of ways to partition 5 into three (possibly empty?) subsets, which is


    for all p+q+r = 5 (and possibly with the requirement that p,q,r are strictyl positive.

    To show that your answer is wrong, I think your reasoning that the number of ways to put 5 distinct objects into 1 box is 5 choose 1, ie 5, when obviously it is 1. Also there are many ways of putting 5 balls into 100 boxes, and 5 choose 100 is define to be 0.
  8. Feb 27, 2004 #7


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    There are 5 possible groupings:
    0 0 5 - 1
    0 1 4 - 5
    0 2 3 - 10
    1 1 3 - 10
    1 2 2 - 20

    For a total of 46 possibilities. I don't see a more general formula for this.
  9. Feb 27, 2004 #8

    matt grime

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    The answer is the partition number of (5,3) for want of a better term which is what you've worked out, NateG, and does not follow the formula I wrote just there. Anyway, it isn't 5 choose 3. Is

    getting better? I think so.

    Last edited: Feb 27, 2004
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