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.
The first method indeed over counts. For example, consider the possibility of only one flavor in the box. The first method would tell you how to choose $3$ flavors from $14$ then see how to arrange these flavors in the box. You could have $12$ of flavor #1, $0$ of flavor #2, and $0$ of flavor #3. But this method also separately counts $12$ of flavor #1, $0$ of flavor #2, and $0$ of flavor #4, for example, which is the same as what we had before. So indeed the first method over counts. This looks to be what you were saying in the question body, but here it is in more detail.
When you physically break the problem into cases like you do in your second method, you remove this over counting that was going on in the first method because you're enforcing the rule that to be considered a part of the case with $3$ flavors in the box, there can't be a flavor represented with $0$ donuts in the box.
All in all, your second approach is the way to go for the combinatorics answer.
However, this isn't the way to find the probability, as you are now mistaking each kind of box as being equally likely to be possible. The probability of having a box of all flavor #1 donuts is not the same as having a box with $11$ flavor #1s and $1$ flavor #2. Think about coin flips: if you flip a coin twice, is the chances of two heads and one head one tails equally likely? To find the probability, take note that this follows a Multinomial Distribution and see if you can go from there.
The probability of this happening is extremely small, as shown by this R script. It generates various possible boxes and checks if the count of different flavors in the box is less than $3$.
total = 0
for (y in 1:10000000){
x = rmultinom(1,12,c(1/14,1/14,1/14,1/14,1/14,
1/14,1/14,1/14,1/14,1/14,1/14,1/14,1/14,1/14))
x <- c(x)
count = 14
for(i in x){
if(i==0){
count = count -1
}
}
if(count <= 3){total = total + 1}
}
sprintf("%.20f", total / 10000000)
One run of this code produced that a box only has at most three flavors $27$ times out of $10,000,000$ boxes.
Best Answer
You are actually asked to find the cardinality of the set:$$\left\{(a_1,\dots, a_{20})\in\mathbb Z_{\geq0}^{20}\,\middle|\, \sum_{i=1}^{20}a_i=12\right\}$$ Here $a_i$ stands for the number of selected donuts that are of kind $i\in\{1,\dots,20\}$.
This can be done by application of stars and bars and indeed leads to outcome $\binom{12+19}{19}=\binom{31}{19}$.
It seems that your "lines" agree with "bars" and your "balls" with "stars", and that your thinking about this is correct.
If there would be less than a dozen donuts of some kind, e.g. there are $10$ of kind $5$ then extra condition $a_5\leq10$ would arise (making things more complex).