Choosing $12$ donuts from $20$ kinds of donuts 
original question: A donut shop offers $20$ kinds of donuts. Assuming
  that there are at least a dozen of each kind as you enter the shop, in
  how many ways can you select a dozen donuts?

The correct answer is: ${}^{31}C_{19}$
I think we can assume that there are $20$ boxes and $12$ balls so that there are $19$ lines and It means choosing $19$ or $12$ from $31$. Is what I think correct?
In addition, I wonder the original question said 'assuming at least a dozen of each kind as you enter the shop'.
 A: 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). 
A: 
It is said to assume that there are at least a dozen donuts of each kind to include the possibility of selecting all 12 donuts of only 1 kind.  

Proceeding by the traditional counting method, assume that:
you select a1 donuts of first kind, a2 of second kind,.... a20 of twentieth kind where
ai$\geq0$ for each i from 1 to 20. 
Since, in total there are only 12 to be chosen, we have $$a1+a2+...+a20=12$$    

Number of integral solutions of a1+a2+...+ar=n is C(n+r-1,r-1) 

Now, you have to find out integral solutions of this equation, which will be $$C((12+20-1),(20-1))$$ =$$C(31,19)$$ 
