(Tom Snowdon Solution to 5.4 54) |
|||
| Line 1: | Line 1: | ||
| − | 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)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) | ||
| Line 13: | Line 13: | ||
TSnowdon@purdue.edu | TSnowdon@purdue.edu | ||
| + | --[[User:Tsnowdon|Tsnowdon]] 23:03, 27 September 2008 (UTC) | ||
Revision as of 18:03, 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 --Tsnowdon 23:03, 27 September 2008 (UTC)
