# How to Count - 3 balls in 3 boxes

I've been spinning my wheels on this for an embarrassing amount of time. This is not for an assignment, I'm just studying.

Three balls are placed at random in three boxes. There is no restriction on the number of balls per box; list the 27 possible outcomes of this experiment.

I could only come up with 10 where order of placing the balls into the boxes doesn't matter.

{(0,0,3), (0,1,2), (0,2,1), (0,3,0), (1,0,2), (1,1,1), (1,2,0), (2,0,1), (2,1,0), (3,0,0)}

What am I doing wrong?

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The problem wants you to treat the three balls as distinct. (I realize this isn't obvious from the statement. It's the only way to get 27 as the answer.) – Qiaochu Yuan Dec 10 '10 at 1:32
Now I have the correct number; thank you. When I considered distinct I also included the order that balls are put into boxes, and the number for that was much larger than 27. However only considering distinctness without respect to order is perfect. – Coltin Dec 10 '10 at 1:50

Using the stars-and-bars technique, for your interpretation of the problem, there are in fact ${3+3-1\choose 3}=10$ possible outcomes.
As Qiaochu Yuan said in his comment, 27 is the result when the three balls are distinguishable (and I agree with him that it is not clear in the problem statement)—for each of the 3 balls, there are 3 possible locations, so $3^3=27$ possible outcomes.