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 \$\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)

( probably related to this thread
http://www.mathhelpforum.com/math-he...subgroups.html )
• Nov 14th 2008, 04:25 AM
Plato
\$\displaystyle {{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 \$\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).
• Nov 14th 2008, 06:42 AM
Plato
Allowing for groups to be empty: \$\displaystyle {{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 \$\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.

Thanks!