(New page: 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 ...) |
(No difference)
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Revision as of 20:36, 24 September 2008
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.
