How many ways of moving $6$ wine bottles from the cellar to the fridge? I have a seemingly easy question about combinatorics but I seem to be missing something for the correct answer. The questions is as follows:  

You are organizing a party and your guests want to drink wine. You
  have 3 different brands in your cellar. Per brand you have 12 bottles.
  In preparation of the party you want to move some bottles to your
  fridge. However your fridge only fits 6 bottles. How many
  possibilities do you have to choose 6 bottles to put in your fridge?

The answer is given as 28 possibilities but I can't seem to figure out why that's the solution.
I tried so many different approaches already but I think I'm missing a catch somewhere.
Any help that would even lead me in the right direction would be appreciated
 A: You should count the number of solution of $i+j+k=6$ where $0 \leqslant i,j,k \leqslant 6$ that is $\binom{6+3-1}{3-1}$.
A: Consider the amount of solutions to
$z+y+x=6,\quad x,y,z\in \mathbb{N}$,
where $x,y,z$ present the amount of different wine bottles.
The generalization $x_1+x_2+\ldots+x_n=N$, is known as stars and bars problem. See this reference https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics), or other related questions on this site.
A: I don’t think it’s very easy.
There are 3 ways of choosing only one brand (choose a brand, take 6 bottles).
There are 15 ways of choosing 2 brands:
3 ways of choosing which brand not to include, multiplied by 5 ways of choosing 6 bottles from the other two brands (since you can take 1-5 of one brand, and 6 minus that for the other).
There are 10 ways of choosing 6 bottles if you include all three brands:
You have to have at least one bottle from each, but you can choose the other three bottles without constraint, so for this sub problem:
3 ways of choosing all three bottles from the same brand, 1 way of choosing all three brands (1 bottle of each), and 6 ways of choosing the three bottles from two brands: 3 ways to choose which not to include, then two choices for which to take 2 of.
Since these are mutually exclusive, there wasn’t any double counting, so there are a total of $3+15+10=28$ ways of choosing the 6 bottles.
