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There are 3 bags and 5 different marbles. In how many ways can the marbles be put into the bags?

(disclosure - the question is one of many in a teacher prep study guide. I am taking the qualification exam for teaching middle school math next month.)

The answer and explanation for the above was given as 3^5 = 243.

Earlier on, the book offered the distinction between say "3 kids of a group of ten are being chosen to play...." and the more specific "3 kids of 10 are being chosen for pitcher/catcher/first base" - the former being a lower number since it's not for specific positions.

With the marble bags, it seems the author counts all in bag one separate from all in bag two, or bag three. I'm okay with a wrong answer, but asking - given a test environment where I'm not going to be able to ask for any clarification, what in the wording above leads to that conclusion that the bags must be treated differently? Note - had the question said "3 different colored bags", I'd have been satisfied with the answer.

Edit - I've passed the exam. Remarkably, this is the most unsatisfying way I've ever had an exam reported, simply pass/fail. One would think especially for math, that the test taker would like a result with precision, even if it's not offered to prospective employers.

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It is difficult, there are unwritten conventions about the formal interpretation of informal words. The conventions are (usually) clear in a particular classroom. But when one is making up problems for a widely taken test, such as a contest, it is useful to err on the side of ponderous explicitness. – André Nicolas Feb 12 '13 at 19:53
up vote 2 down vote accepted

There are 3 bags and 5 different marbles. In how many ways can the marbles be put into the bags?

One of the more common unstated assumptions mentioned by André Nicolas in the comments is that containers, unlike the objects put into them, are distinguishable unless otherwise stated. In a problem asking for the number of ways to put marbles into bags or boxes, the marbles may be distinguishable or not, and you’re usually explicitly told which is the case. (If you’re not told, the odds probably favor the indistinguishable interpretation.) For the containers, however, the default seems to be distinguishable containers. Thus, you typically get either a simple $k$-fold choice made $n$ times (with $n$ distinguishable marbles and $k$ distinguishable bags) or a stars-and-bars problem (with $n$ indistinguishable marbles and $k$ distinguishable bags).

There is a good reason for this: the corresponding problems with indistinguishable bags are significantly harder. The closed forms $k^n$ and $\binom{n+k-1}{k-1}$ for the two problems above are very and relatively straightforward, respectively. The answer to the problem with distinguishable marbles and indistinguishable bags is $n\brace k$, a Stirling number of the second kind, which has no nice closed form. And if both marbles and bags are indistinguishable, the answer is given by one of the partition functions $q(n,k)$ and $P(n,k)$, depending on whether or not we allow bags to be empty; these also lack nice closed forms.

That said, it would certainly have been better for the proposer to have specified different bags as well as different marbles.

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"If you’re not told, the odds probably favor the indistinguishable interpretation." - Thanks again for the detailed response. I'm hopeful the exam itself will have a bit more clarity, but I'll go with this advice, especially knowing now that my interpretation was actually more complex. – JoeTaxpayer Feb 16 '13 at 15:05
@JoeTaxpayer: You’re welcome; good luck with the exam. – Brian M. Scott Feb 16 '13 at 22:58
Update - the exam was today. The questions that fell into this category were few, one was a simple factorial question along the lines of "5 kids all sitting in front row in five seats, how many different ways can they arrange themselves?" The other question offered 4 answers as equations, and it was simple matter to eliminate the incorrect ones. Glad I was prepared for what came up. (score will arrive in one month to close this saga) – JoeTaxpayer Mar 2 '13 at 20:08
@JoeTaxpayer: Sounds like it went well; glad to hear it! – Brian M. Scott Mar 2 '13 at 20:12

Hint: Think about the balls. In how many bags can one ball be put in? How many balls are there?

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If the three bags are sitting on the table and I rotate the contents, there's still one ball in the one bag. Let me add - if I were loading 3 marbles into 3 bags, just one per bag, there would be two answers depending on if the bags are uniquely identified. If they are three brown bags, one answer, but three numbered or different colored bags, different answer. – JoeTaxpayer Feb 12 '13 at 19:51
nevertheless, there is no restrictions such as you implied in this question. – ciceksiz kakarot Feb 12 '13 at 19:58
The author's solution treats them as unique. But the question avoids the distinction. – JoeTaxpayer Feb 12 '13 at 20:03

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