# partitioning in a number of groups

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• Nov 14th 2008, 02:34 AM
Simo
partitioning in a number of groups
Someone know how to compute in how many ways is it possible to partition a collection of $M$ indistinguishable elements in $I$ groups (evidently with at least one element for each groups)???
The groups are distinguishable; I mean, if $M=3$ and $I=2$ the solution is 2 combinations (1,2) and (2,1) (whereas (3,0) and (0,3) are not allowed since a group is empty)

( probably related to this thread
http://www.mathhelpforum.com/math-he...subgroups.html )
• Nov 14th 2008, 04:25 AM
Plato
${{M-1}\choose{I-1}}$
• Nov 14th 2008, 05:22 AM
Simo
Thank you very much...
and admitting also an indefinite number of "empty" groups within the total number of groups $I$? so, in the same example of above, if $M=3$ and $I=2$ the solution is 4 combinations: (3,0), (2,1), (1,2), (0,3).
• Nov 14th 2008, 06:42 AM
Plato
Allowing for groups to be empty: ${{M+I-1}\choose{M}}$ .
• Nov 18th 2008, 01:45 AM
Simo
the last question...I hope (Nod)
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 $M$ distinguishable elements in $I$ distinguishable groups, with and without admitting for empty groups.
For example, if $M=2$ and $I=2$:
- without admitting for empty groups: $(obj_1, obj_2), (obj_2, obj_1) == 2$ combinations.
- admitting empty groups: $(obj_1, obj_2), (obj_2, obj_1), (obj_1 \& obj_2, nothing), (nothing, obj_1 \& obj_2) == 4$ combinations.

Thanks!