(New page: There are 5 possible groupings: 0 0 5 - 1 0 1 4 - 5 0 2 3 - 10 1 1 3 - 10 1 2 2 - 20 For a total of 46 possibilities. I don't see a more general formula for this. Because [A][B][C] and [B...) |
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For a total of 46 possibilities. I don't see a more general formula for this. Because [A][B][C] and [B][A][C] are same, the boxes are indistinguishable that proves. | For a total of 46 possibilities. I don't see a more general formula for this. Because [A][B][C] and [B][A][C] are same, the boxes are indistinguishable that proves. | ||
Revision as of 18:47, 4 October 2008
There are 5 possible groupings: 0 0 5 - 1 0 1 4 - 5 0 2 3 - 10 1 1 3 - 10 1 2 2 - 20
For a total of 46 possibilities. I don't see a more general formula for this. Because [A][B][C] and [B][A][C] are same, the boxes are indistinguishable that proves.
As a result, 5 orders to indicate 5 indistinguishable objects into 3 indistinguishable boxes.
