[Math] How many ways can you get a bag of $12$ donuts if a shop sells $30$ kinds of donuts

combinationscombinatoricsdiscrete mathematics

A donut shop sells $30$ kinds of donuts, and there are at least $12$ donuts of each kind. How many ways can you:

  1. Get a bag of $12$ donuts?
  2. Get a bag of $12$ donuts if you want at least $3$ glazed donuts and at least $4$ chocolate donuts?
  3. Get a bag of $12$ donuts if you want exactly $3$ glazed donuts and exactly $4$ chocolate donuts?

For the first one I am thinking it is $C(30+12-1,12)=C(41,12)=7898654920$. This seems like a big number, but I think it's correct. I am not sure how I'd go about solving the second and third, though.

EDIT:
For 2, perhaps I do the following? I can choose $3$ glazed and $4$ chocolate in $\frac{7!}{3!4!}=35$ ways and that leaves me to choose $5$ donuts from the remaining $30$ kinds, so $C(30+5-1,5)=278256$. Then multiplying together I would get $9738960$ ways. Is this correct?

Best Answer

Problem 1 is standard Stars and Bars. You solved it correctly. The number of ways is $\binom{30+12-1}{12}$, or equivalently $\binom{30+12-1}{30-1}$.

For Problem 2, I think we are to assume that glazed is one of the $30$ kinds, and that all glazed doughnuts (from that store) are identical. Ditto for chocolate.

So if we want at least $3$ glazed and at least $4$ chocolate, put these in the bag. We need to get a further $5$ from the $30$ kinds. Solve as in the first problem. The number of ways is $\binom{30+5-1}{5}$. Note that your answer contained this number. The $\binom{7}{3}$ that you multiplied by should not be there.

For Problem 3, things are pretty much the same, except we must choose $5$ doughnuts from the $28$ kinds remaining, since we don't want any more glazed or chocolate.

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