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)
