Revision as of 16:50, 6 October 2008 by Lee462 (Talk)

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!
-- by lee462 10:48, 5 October 2008 (UTC)

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