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Thread: partitioning in a number of groups

  1. #1
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    partitioning in a number of groups

    Someone know how to compute in how many ways is it possible to partition a collection of $\displaystyle M$ indistinguishable elements in $\displaystyle I$ groups (evidently with at least one element for each groups)???
    The groups are distinguishable; I mean, if $\displaystyle M=3$ and $\displaystyle I=2$ the solution is 2 combinations (1,2) and (2,1) (whereas (3,0) and (0,3) are not allowed since a group is empty)

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  2. #2
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    $\displaystyle {{M-1}\choose{I-1}}$
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  3. #3
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    Thank you very much...
    and admitting also an indefinite number of "empty" groups within the total number of groups $\displaystyle I$? so, in the same example of above, if $\displaystyle M=3$ and $\displaystyle I=2$ the solution is 4 combinations: (3,0), (2,1), (1,2), (0,3).
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  4. #4
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    Allowing for groups to be empty: $\displaystyle {{M+I-1}\choose{M}}$ .
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  5. #5
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    the last question...I hope
    is there also a simple form for to give an answer to the previous two posts in the case of distinguishable elements?
    So, the number of combinations of $\displaystyle M$ distinguishable elements in $\displaystyle I$ distinguishable groups, with and without admitting for empty groups.
    For example, if $\displaystyle M=2$ and $\displaystyle I=2$:
    - without admitting for empty groups: $\displaystyle (obj_1, obj_2), (obj_2, obj_1) == 2$ combinations.
    - admitting empty groups: $\displaystyle (obj_1, obj_2), (obj_2, obj_1), (obj_1 \& obj_2, nothing), (nothing, obj_1 \& obj_2) == 4$ combinations.

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