How many ways can we select 3 elements from the multiset G = {A, B, B, C, C, C}? The problem is to determine how many distinct sets of 3 elements taken from multiset G.
So we have the following 6 ways:
(A, B, C)
(A, B, B)
(A, C, C)
(B, C, C)
(B, B, C)
(C, C, C)
*Note that element (A) when chosen, will only appear 1 time in the selection sub-set, because it only has 1 element (A).
But what is the formula to find this amount?
The problem is actually finding a generic formula for any case, as shown in the figure below:

Where:
a = number of elementos A,
b = number of elementos B,
c = number of elementos C,   and so on...
What is the formula or method to find how many different ways we can form a set of (p) elements, selecting (p) elements from a multiset G, with (t) types of elements with certain repetitions?
 A: coefficient of $x^p$ in $(1+x^1+x^2+x^3........x^a)(1+x^1+x^2.......x^b)(1+x^1+x^2........x^c).....................$ is your required answer.
As your question just depends on how many A or B or C we took. For example if you  take 3 A then use $x^3$ from first set, take 2 B then use $x^2$ from second set and so on.
This of course is not easy to calculate generally, but if it is given that a,b,c,d.......x (all of them)$\ge$p, then it is a very general result giving value ${t+p-1 \choose p}.$
A: You could also use inclusion/exclusion:
You are looking for $$x_A+x_B+x_C = 3, 0\le x_A \le 1, 0\le x_B \le 2, 0\le x_C \le 3$$
The total number of solutions with no upper bounds is $$\dbinom{3+3-1}{3-1}$$
Suppose the upper bound for $x_A$ was violated. This implies $x_A \ge 2$. So, subtract 2 from the RHS: $$x_A^\prime+x_B+x_C = 3-2 = 1$$ So, the upper bound is violated in $$\dbinom{1+3-1}{3-1}$$ ways.
Next, suppose the upper bound for $x_B$ was violated. This implies $x_B \ge 3$. So, subtract 3 from the RHS: $$x_A+x_B^\prime+x_C = 3-3 = 0$$ So, the upper bound is violated in $$\dbinom{0+3-1}{3-1}$$ ways.
There are no ways to violate more than one upper bound at the same time, so the total is $$\dbinom{5}{2}-\dbinom{3}{2}-\dbinom{2}{2} = 10-3-1=6$$

For a general formula: instead of using $a,b,c,\ldots, x$, I will use $x_1,x_2,x_3, \ldots, x_t$ so that the indexes are able to be iterated. Let $[t] = \{1,2,\ldots, t\}$. 
$$\sum_{A \subseteq [t]}(-1)^{|A|}\dbinom{p+t-1-|A|-\displaystyle \sum_{a \in A}x_a}{t-1}$$
Note: for this formula, we are defining $\dbinom{n}{r}$ to take the value of $0$ for all $n<r$.
