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| 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 |
| Line 10: | Line 10: | ||
As a result, | As a result, | ||
| − | 5 orders to indicate 5 indistinguishable objects into 3 indistinguishable boxes. | + | 5 orders to indicate 5 indistinguishable objects into 3 indistinguishable boxes.--[[User:Kim297|Kim297]] 21:56, 6 October 2008 (UTC) |
| + | |||
| + | |||
| + | --- | ||
| + | '''grader's comment''' : good job. I also got the same answer for the question! <Br> | ||
| + | -- by [[User:lee462|lee462]] 10:48, 5 October 2008 (UTC) | ||
Latest revision as of 18:30, 6 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.--Kim297 21:56, 6 October 2008 (UTC)
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grader's comment : good job. I also got the same answer for the question!
-- by lee462 10:48, 5 October 2008 (UTC)
