Revision as of 19:36, 24 September 2008 by Mkburges (Talk)

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Yeah, so, for putting items into indistinguishable boxes, really has not theorem in our knowledge, as of yet. So the easiest way to do this is to directly express them and then count.

5 objects into 3 groups: {ABCDE} {ABCD,E} {ABC,DE} {ABC,D,E} {AB,CD,E}

Hence 5 different groups.

--Michael Burgess

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood