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There are two situations.The first is a bakery which has three type of doughnuts, {6*chocolate , 6*cinnamon, 3*plain}. How many options do they have for a box of 12 doughnuts?

The second question is what is the number of ways 10 people can be placed in 6 rooms? My actual question is not the answer for those questions but rather the first step. They are both problems in the chapter of Inclusion-exclusion. The first step is to count the total possible ways that each can occur without restrictions.

In my mind in both situations I can think of having a multiset with 3 or 6 elements respectively with infinite amounts of each. The second problem was worked out in class and the total without restrictions was 6^10. Clearly, this makes sense because each person could choose any room.

The professor made a point that $$\binom{10+(6-1)}{10}$$ was not the answer. Even though it seems to me that this would be valid. Couldn't one think of the rooms as a set of "infinite" rooms which you are making 10 choices from, one for each person? I think I'm having a hard time understanding where the order comes from that turns the total into a permutation rather than a combination.

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up vote 2 down vote accepted

You can think of the first problem as distributing $12$ choices amongst $3$ kinds of doughnuts. This is like distributing $12$ identical marbles amongst $3$ labelled boxes: it doesn’t matter which two choices go in the plain doughnut box, so to speak.

You cannot think of the second problem as distributing $10$ identical marbles amongst $6$ boxes, however, because the ‘marbles’ (i.e., the people) are not identical: it matters which two people you put in Room $3$, for instance.

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