Rolling five $6$-sided dice, where (as an example) $3$, $2$ and $1$ faces of each die are equivalent Let's say we have a $6$ sided die. $3$ of the sides have the value $A$, $2$ have the value $B$ and $1$ has the value $C$.
If we take $5$ of those, roll them together and look at the possible values, how would I go about finding the probability for specific combinations? Examples of combinations where X denotes any value could be:


*

*$AAXXX$

*$AAAXX$

*$AABBX$

*$ABCXX$

*$4$ of a kind
Is there a general way to figure the probability of obtaining a combination from a roll of $5$ dice? Both for a die face distribution $3A2B1C$ and possibly other alternate distributions like $3A1B1C1D$. 
Thank you for any help you can give. I'm sorry if the question is a bit bare bones on my end but I was a bit unsure on exactly how to approach this.
 A: Someone else said use the multinomial distribution, but to be a little more clear:
If you have values $v_1,v_2,\ldots, v_k$ with multiplicity numbers $m_1,m_2,\ldots, m_k$ with $\sum_{i=1}^k m_i = M$, then the probability that a single fair die will roll value $v_i$ is $\dfrac{m_i}{M}$.
So, if you roll $n$ dice and you want to know the probability of getting $a_1\cdot v_1, a_2\cdot v_2,\ldots, a_k\cdot v_k$, it is given by:
$$\dbinom{n}{a_1,a_2,\cdots a_k}\prod_{i=1}^k\left(\dfrac{m_i}{M}\right)^{a_i}$$
In your example where you have $A\cdot 3, B\cdot 2, C\cdot 1$, the probability of rolling $A$ is $\dfrac{3}{3+2+1} = \dfrac{1}{2}$, the probability of rolling $B$ is $\dfrac{2}{3+2+1} = \dfrac{1}{3}$, and the probability of rolling $C$ is $\dfrac{1}{3+2+1} = \dfrac{1}{6}$.
Since you are rolling five dice, the probability of getting, say, $3\cdot A, 2\cdot B$ would be:
$$\dbinom{5}{3,2,0}\left(\dfrac{1}{2}\right)^3\left(\dfrac{1}{3}\right)^2\left(\dfrac{1}{6}\right)^0 = \dfrac{5!}{3!2!0!}\left(\dfrac{1}{8}\right)\left(\dfrac{1}{9}\right) = \dfrac{5}{36}$$
Extrapolating over all possible outcomes:
$$\begin{array}{c|c}\text{Outcome} & \text{Probability} \\ \hline \{A\cdot 5\} & \dfrac{1}{32} \\ \{A\cdot 4, B\cdot 1\} & \dfrac{5}{48} \\ \{A\cdot 4, C\cdot 1\} & \dfrac{5}{96} \\ \{A\cdot 3, B\cdot 2\} & \dfrac{5}{36} \\ \{A\cdot 3,B\cdot 1, C\cdot 1\} & \dfrac{5}{36} \\ \{A\cdot 3, C\cdot 2\} & \dfrac{5}{144} \\ \{A\cdot 2, B\cdot 3\} & \dfrac{5}{54} \\ \{A\cdot 2,B\cdot 2, C\cdot 1\} & \dfrac{5}{36} \\ \{A\cdot 2, B\cdot 1, C\cdot 2\} & \dfrac{5}{72} \\ \{A\cdot 2, C\cdot 3\} & \dfrac{5}{432} \\ \{A\cdot 1, B\cdot 4\} & \dfrac{5}{162} \\ \{A\cdot 1, B\cdot 3, C\cdot 1\} & \dfrac{5}{81} \\ \{A\cdot 1, B\cdot 2, C\cdot 2\} & \dfrac{5}{108} \\ \{A\cdot 1, B\cdot 1, C\cdot 3\} & \dfrac{5}{324} \\ \{A\cdot 1,C\cdot 4\} & \dfrac{5}{2592} \\ \{B\cdot 5\} & \dfrac{1}{243} \\ \{B\cdot 4, C\cdot 1\} & \dfrac{5}{486} \\ \{B\cdot 3, C\cdot 2\} & \dfrac{5}{486} \\ \{B\cdot 2, C\cdot 3\} & \dfrac{5}{972} \\ \{B\cdot 1, C\cdot 4\} & \dfrac{5}{3888} \\ \{C\cdot 5\} & \dfrac{1}{7776}\end{array}$$
Note: If you add up the probabilities for every possible outcome I listed, you will get 1 (100% probability that you wind up with one of those outcomes). This is known as the Law of Total Probability and shows that I did correctly describe the entire probability space of rolling five dice.
Since each outcome is disjoint, for any case you are trying to build (like at least two A's), find all outcomes that have at least two A's, and add their probabilities. For 4-of-a-kind, find every outcome with exactly 4 of a kind and add up their probabilities.
Also note: I strictly interpreted your question to be a question of dice rolled in any order. Rolling AAABB is the same as rolling ABBAA is the same as rolling ABABA, etc. in order.
