Ball and urn problem Let's say there are three types of balls, labeled $A$, $B$, and $C$, in an urn filled with infinite balls. The probability of drawing $A$ is $0.1586$, the probability of drawing $B$ is $0.81859$, and the probability of drawing $C$ is $0.02275$.
If you draw $6$ balls, what's the probability that you drew $3 A$s, $2 B$s, and $1 C$?
I know what the answer is but I don't know how to set up the problem appropriately. It's not just a simple counting problem because the balls have weighted probabilities... Any push in the right direction is appreciated!

Any sequence of the sort $ABACAB$ has the same probability to occur. So it comes to finding how many distinct words there are having $3$ times an $A$, $2$ times a $B$ and $1$ time a $C$. 
You could start with $6$ open spots and then placing the $A$'s. That gives $\binom{6}{3}$ possibilities. 
Then place the $B$'s. That gives $\binom{3}{2}$ and finally place $C$ on the single spot that is left. 
There are $\binom{6}{3}\times\binom{3}{2}=60$ possibilities so the corresponding probability is $60\times p_A^3p_B^2p_C$ 
