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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.