(New page: 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 o...) |
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TSnowdon@purdue.edu | TSnowdon@purdue.edu | ||
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| + | --[[User:Tsnowdon|Tsnowdon]] 23:06, 27 September 2008 (UTC) | ||
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| + | Tom, well done on getting the correct answer! I've also come to believe that this is the most clever way of solving this problem. It is shorter/easier to realize that all of the possible outcomes are simply the ways for three numbers to sum to 5, while keeping in mind that the boxes are indistinguishable of course, than listing all of the possibilities like I did. A+<br><br> | ||
| + | --[[User:Aoser|Aoser]] 12:41, 3 October 2008 (UTC)<br> | ||
| + | Aoser@purdue.edu | ||
Latest revision as of 07:42, 3 October 2008
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 23:06, 27 September 2008 (UTC)
Tom, well done on getting the correct answer! I've also come to believe that this is the most clever way of solving this problem. It is shorter/easier to realize that all of the possible outcomes are simply the ways for three numbers to sum to 5, while keeping in mind that the boxes are indistinguishable of course, than listing all of the possibilities like I did. A+
--Aoser 12:41, 3 October 2008 (UTC)
Aoser@purdue.edu
