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Math Help - Probability of each combination with repetition

  1. #1
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    Probability of each combination with repetition

    Hi guys,

    It looks like trivial problem but I cannot solve it.

    I have M baskets and N balls. I can choose one of the baskets randomly and put a ball into the chosen basket.

    I'm just interested in combination. So in three baskets with 5 balls,

    (1) (1) (3)

    (1) (3) (1)

    (3) (1) (1)

    above combinations are same for me.

    I just want to know the probability of each combinations. I know that all possible combination can be represented by combination with repetition.

    But I do not know that how many cases of each combinations are existed.

    Any helps are appreciated!!!

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by ychang View Post
    I have M baskets and N balls. I can choose one of the baskets randomly and put a ball into the chosen basket.
    (1) (1) (3); (1) (3) (1); (3) (1) (1)
    (the)above combinations are same for me.
    This a poorly worded question. Read it again considering my highlights.
    Can a basket be empty? If not then there are just two possible answers for you example:
    5=3+1+1 or 5=2+2+1.

    If baskets can be empty you have: 5=5; =4+1; 5=3+1+1; or 5=2+2+1

    So the answer depends upon what you mean.
    Please respond with a clarification.
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  3. #3
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    Thanks Plato,

    Yes, basket can be empty. And I know that \frac{(M+N-1)!}{(M-1)!N!}@ ways are possible. Can I know that how many times each possible combinations are occured?

    Thanks again!
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  4. #4
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    Quote Originally Posted by ychang View Post
    Yes, basket can be empty.
    Then I think that you asking about the number of partitions of an integer in k or fewer summands.
    Example: Say that there are seven balls and four baskets, then the answer is eleven.
    7; 6+1, 5+2, 4+3; 5+1+1, 4+2+1, 3+2+2, 3+3+1; 4+1+1+1, 3+2+1+1, 2+2+2+1.

    The formula is not easy. In fact it is given as a recursive function.
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