| Line 13: | Line 13: | ||
| − | + | --- | |
| − | grader's comment : good job. I also got the same answer for the question! | + | '''grader's comment''' : good job. I also got the same answer for the question! <Br><BR> |
--[[User:lee462|lee462]] 10:48, 5 October 2008 (UTC) | --[[User:lee462|lee462]] 10:48, 5 October 2008 (UTC) | ||
Revision as of 16:50, 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.
---
grader's comment : good job. I also got the same answer for the question!
--lee462 10:48, 5 October 2008 (UTC)
