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--[[User:Kduhon|Kduhon]] 09:52, 24 September 2008 (UTC) | --[[User:Kduhon|Kduhon]] 09:52, 24 September 2008 (UTC) | ||
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| + | I believe this is a correct answer. There are no more than 5 answers because the boxes are indistinguishable and thus 221 is the same as 122 and 121, etc. Because the book says there is no formula for this method of indistinguishable boxes, i believe this is the most efficient way of doing this because we were given such small numbers. Any longer of a method would have been marginally not worth the time for this. It is a clever solution, well done. | ||
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| + | -[[Aifrank_MA375Fall2008walther]] | ||
Revision as of 18:09, 5 October 2008
OK, so we need to put 5 indistinguishable objects into 3 indistinguishable boxes.
The easiest way to do this is to just list the different possibilities.
5 0 0 4 1 0 3 2 0 3 1 1 2 2 1
Each column is a box and each row is a possibility. Notice that I did not label the boxes, this is because they are indistinguishable. This makes it so I do not have to list 5-0-0, 0-5-0, and 0-0-5 as separate possibilities. They are all the same. Also, options like 4-1-0 only has one possibility because the objects are also indistinguishable so it doesn't matter which object is separate from the others.
Add up these possibilities and we get 5 ways.
--Kduhon 09:52, 24 September 2008 (UTC)
I believe this is a correct answer. There are no more than 5 answers because the boxes are indistinguishable and thus 221 is the same as 122 and 121, etc. Because the book says there is no formula for this method of indistinguishable boxes, i believe this is the most efficient way of doing this because we were given such small numbers. Any longer of a method would have been marginally not worth the time for this. It is a clever solution, well done.
