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 21:03, 5 October 2008
The problem to me is somewhat frustrating to grade due to the fact that the subject of indistinguishable objects into indistinguishable boxes is a shady subject in the text. What happens if the number of ways is too large to physically count out? I wouldn't know as the book didn't explain this to me. From information gathered in the book's illustration to a similar problem (example 11 in 5.5), your answer is tip-top and I would give you full credit. The only discrepancy that Chris and I had was stating the problem, and he already mentioned that. I like the way that you wrote out the reasoning to the first two possibilities that you wrote. It gives the reader a clear understanding of the path that you are headed on.
--Ehanna 22:01, 5 October 2008 (UTC)
