I made three colums, all labeled "Box" to signify the three indistinguishable boxes. Writing out the combinations I found there could be
5 in one
4 in one and 1 in another
3 in one, 1 in another, and 1 in the last
3 in one and 3 in another
2 in one, 1 in another, and 2 in the last
for a total of five solutions.
--Tmsteinh 17:51, 24 September 2008 (UTC)
Great job! This method/solution looks perfect to me :) --Zhao14 08:11, 3 October 2008 (UTC)
