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Writing out the combinations I found there could be | Writing out the combinations I found there could be | ||
| − | 5 in one | + | |
| − | 4 in one and 1 in another | + | 5 in one |
| − | 3 in one, 1 in another, and 1 in the last | + | |
| − | 3 in one and 3 in another | + | 4 in one and 1 in another |
| − | 2 in one, 1 in another, and 2 in the last | + | |
| + | 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. | for a total of five solutions. | ||
--[[User:Tmsteinh|Tmsteinh]] 17:51, 24 September 2008 (UTC) | --[[User:Tmsteinh|Tmsteinh]] 17:51, 24 September 2008 (UTC) | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | Great job! This method/solution looks perfect to me :) | ||
| + | --[[User:Zhao14|Zhao14]] 08:11, 3 October 2008 (UTC) | ||
Latest revision as of 04:11, 3 October 2008
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)
