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It seems it is this problem is easy enough to write out all the possibilities.
 
It seems it is this problem is easy enough to write out all the possibilities.
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--[[User:Jahlborn|Jahlborn]] 22:34, 28 September 2008 (UTC)
 
--[[User:Jahlborn|Jahlborn]] 22:34, 28 September 2008 (UTC)
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This is the same answer that i got along with most others.  It is clear and easy to understand.  I would have written out the question so it is easy to follow if i do not have a book in front of me.  Good job! I give you an A in my book.
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--[[:ccuriel_MA375Fall2008walther|ccuriel]]

Revision as of 15:03, 5 October 2008

It seems it is this problem is easy enough to write out all the possibilities.

poss 1. (5)(0)(0) This stands for all five items in any of the three boxes because they are indistinguishable.

poss 2. (4)(1)(0) This stands for four items in one of the three boxes and the remaining item in one of the remaining two boxes.

poss 3. (3)(2)(0) Same reasoning.

poss 4. (3)(1)(1)

poss 5. (2)(2)(1)


--Jahlborn 22:34, 28 September 2008 (UTC)


This is the same answer that i got along with most others. It is clear and easy to understand. I would have written out the question so it is easy to follow if i do not have a book in front of me. Good job! I give you an A in my book.

--ccuriel

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett