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| − | --[[User:Tsnowdon|Tsnowdon]] 23:03, 27 September 2008 (UTC)There are five "skeleton" outcomes. These are [5][0][0], [4][1][0], [3][2][0], [3][1][1], [2][2][1]. | + | --[[User:Tsnowdon|Tsnowdon]] 23:03, 27 September 2008 (UTC) |
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| + | There are five "skeleton" outcomes. These are [5][0][0], [4][1][0], [3][2][0], [3][1][1], [2][2][1]. | ||
(where [#] is a box with # elts) | (where [#] is a box with # elts) | ||
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TSnowdon@purdue.edu | TSnowdon@purdue.edu | ||
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Latest revision as of 18:04, 27 September 2008
--Tsnowdon 23:03, 27 September 2008 (UTC)
There are five "skeleton" outcomes. These are [5][0][0], [4][1][0], [3][2][0], [3][1][1], [2][2][1]. (where [#] is a box with # elts)
But the elements are indistinguishable, so there is only one way to obtain [x][y][z] (with x+y+z =5 obviously). Also the boxes are indistinguishable, so we must consider [x][y][z] and [y][z][x] to be the same outcome.
Thus the 5 "skeleton" outcomes are the only outcomes.
i.e. there are 5 ways to distribute 5 indistinguishable objects into 3 indistinguishable boxes.
By Tom Snowdon,
TSnowdon@purdue.edu
