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Where a group can consist of 1 or more items, groups don't have to be equally sized and items can be duplicates. Example - Choose 3 groups:

Items: 1 2 2 3

Groups:

(1) (2 2) (3)

(1 2) (2) (3)

(3 1) (2) (2)

(3 2) (2) (1)

Update Following Rasmus' comment, I see that this is more technically described as finding the possible partitions of a multiset where the number of cells is x.

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    This may be a duplicate of [this question](http://math.stackexchange.com/questions/16890/stirling-numbers-of-the-second-kind-on-multiset?rq=1), assuming that the other question allows multiset parts (i.e., the individual parts making up the partition can contain duplicate elements).2013-06-21

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Note that this corresponds to the partition of $4$ into $3$ nonzero parts (or $2+1+1 = 4$).

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    Not quite, since the items have identities (which may be shared). The general formula will be a mess (some coefficient of something in some generating function).2011-03-06