View source for Yongseok Kim's Solution to 5.5 MA375Fall2008walther - Rhea

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.