My Math Forum Possible exhaustive groupings of N objects?

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 May 10th, 2007, 08:38 AM #1 Newbie   Joined: May 2007 Posts: 2 Thanks: 0 Possible exhaustive groupings of N objects? I have a question: If you have N objects, what is the possible number of ways that you can exhaustively group them (i.e. group them without having any objects left over?) For example: for N = 2; there are 2 possible groupings: [AB] [A] + [B] for N = 3; there are 5 possible groupings: [ABC] [AB] + [C] [AC] + [B] [BC] + [A] [A] + [B] + [C] for N=4, there are (I think) 15: [ABCD] [ABC]+[D] [ABD]+[C] [ACD]+[B] [BCD]+[A] [AB]+[CD] [AB]+[C]+[D] [A]+[B]+[CD] [BC]+[AD] [BC]+[A]+[D] [B]+[C]+[AD] [AC]+[BD] [A]+[C]+[BD] [AC]+[B]+[D] [A]+[B]+[C]+[D] Any help would be greatly appreciated. Thanks, Ed
 May 10th, 2007, 10:45 AM #2 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Hmm... Is this related to the problem of putting n labeled balls into n indistinguishable boxes? See sequence A000110.
May 10th, 2007, 11:38 AM   #3
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Joined: May 2007

Posts: 2
Thanks: 0

I think you're right. I saw this before, but was thrown by the main title. Looking lower, it does say:
Quote:
 Number of partitions of an n-element set.
which is what I'm looking for. (Slaps forehead) D'oh! Of course it's the same: any empty "boxes" in the answer don't matter:

[ABC] + [] + []
[AB] + [C] + []
[AC] + [B] + []
[BC] + [A] + []
[A] + [B] + [C]

is equivalent to

[ABC]
[AB] + [C]
[AC] + [B]
[BC] + [A]
[A] + [B] + [C]

...at least for calculating the number of possible solutions. I found a nice description is on the WikiPedia: http://en.wikipedia.org/wiki/Partition_of_a_set

Thanks a lot for your help!
-Ed

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