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


*

*Get a bag of $12$ donuts?

*Get a bag of $12$ donuts if you want at least $3$ glazed donuts and at least $4$ chocolate donuts?

*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?

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. 
