| Line 1: | Line 1: | ||
| − | There are 5 possible groupings: | + | There are 5 possible groupings: 0 0 5 - 1 |
| − | 0 0 5 - 1 | + | 0 1 4 - 5 |
| − | 0 1 4 - 5 | + | 0 2 3 - 10 |
| − | 0 2 3 - 10 | + | 1 1 3 - 10 |
| − | 1 1 3 - 10 | + | 1 2 2 - 20 |
| − | 1 2 2 - 20 | + | |
Revision as of 18:48, 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.
