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A: Similarly to everyone else, I just made a table for the possible situations. | A: Similarly to everyone else, I just made a table for the possible situations. | ||
| − | + | 5 - 0 - 0 | |
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| + | 4 - 1 - 0 | ||
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| + | 3 - 2 - 0 | ||
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| + | 2 - 2 - 1 | ||
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| + | 1 - 3 - 1 | ||
These are all '''5''' possible cases. There would be more cases for each entry, such as on the top line (5,0,0) (0,5,0) (0,0,5). But because these are indistinguishable boxes, you only count one of the three cases per possibility. | These are all '''5''' possible cases. There would be more cases for each entry, such as on the top line (5,0,0) (0,5,0) (0,0,5). But because these are indistinguishable boxes, you only count one of the three cases per possibility. | ||
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--[[User:Hschonho|Mike Schonhoff]] 16:09, 24 September 2008 (UTC) | --[[User:Hschonho|Mike Schonhoff]] 16:09, 24 September 2008 (UTC) | ||
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| + | Great work, Mike! This is what I did, so obviously I agree... Since there's no explicit formula, I think this is probably the best way to handle a problem like this, especially since the numbers are small enough to be manageable. | ||
| + | --[[User:Tmsteinh|Tmsteinh]] 09:34, 2 October 2008 (UTC) | ||
Latest revision as of 05:34, 2 October 2008
54. How many ways are there to distribute five indistinguishable objects into three indistinguishable boxes?
A: Similarly to everyone else, I just made a table for the possible situations.
5 - 0 - 0
4 - 1 - 0
3 - 2 - 0
2 - 2 - 1
1 - 3 - 1
These are all 5 possible cases. There would be more cases for each entry, such as on the top line (5,0,0) (0,5,0) (0,0,5). But because these are indistinguishable boxes, you only count one of the three cases per possibility.
The answer is 5.
--Mike Schonhoff 16:09, 24 September 2008 (UTC)
Great work, Mike! This is what I did, so obviously I agree... Since there's no explicit formula, I think this is probably the best way to handle a problem like this, especially since the numbers are small enough to be manageable. --Tmsteinh 09:34, 2 October 2008 (UTC)
