Probability question - combinatorics: 12 tea flavors and 4 customers A tea shoppe serves $12$ different flavors of tea. $4$ customers each order a cup of tea. Assuming that each customer's choice is completely random and is independent of any other customer's selection:
1) what is the probability that $3$ different flavors are selected by the $4$ customers (e.g:: if letters $A-L$ represent flavors, $AABC$)?
2) what is the probability that only $2$ different flavors are selected by these $4$ customers (e,g: $AABB$ or $AAAB$)?
3) what is the probability that $2$ students choose $1$ flavor and the other $2$ choose another flavor (e.g: $AABB$)?
So what is apparent to me is that the total number of possible outcomes is $12^4 = 20736$; the number of possible outcomes where they all choose the same flavor is $(12 P 1) = 12$; the number of possible outcomes where all choose different flavors is $12 P 4 = 11880$, and so the number of outcomes for everything else must be $20736 - 11880 - 12 = 8844$. So the total possible number of outcomes for #1 and #2 above should equal $8844$, but I am not getting this (note that my question 3 is a subset of 2).
I set up #2 as $$\frac{(12 P 2) \cdot 2 \cdot 2}{12^4} = \frac{132 \cdot 2 \cdot 2}{12^4} = \frac{528}{12^4}$$ Note that the factors $2 \cdot 2$ in the numerator are to account for the $3$rd and $4$th customers, who each can choose either $1$ of the $2$ flavors chosen by the first $2$ customers.
I set up #3 as $$\frac{(12 P 3) *3}{12^4} = \frac{1320 \cdot 3}{12^4} = \frac{3960}{12^4}$$  Note that the factor of $3$ in the numerator is to account for the $4$th customer, who can choose any $1$ of the $3$ flavors already chosen by the other $3$ customers.
So the total number of successful outcomes for #1 and #2 are $4488$, instead of $8844$ (just a coincidence that the numbers are reversed or did i do something dyslexic?)
Note: This problem is derived from a much easier problem in a high school algebra II class, where the question was "what is the probability that at least 2 of the customers select the same flavor?" The answer to this is much easier using 1 - the compliment, with the compliment being the probability that all the customers select a different flavor, so answer is $$\frac{20736 - 11880}{20736} = .4271$$
However, I want to show the probability for each of the mutually exclusive possible outcomes: they all select different flavors $$\frac{12 P 4}{12^4}$$ they all select the same flavor $$\frac{12}{12^4}$$; $3$ select the same flavor and $1$ chooses something different; $2$ select one flavor and the other $2$ select another flavor; and finally, $2$ choose the same flavor and the other $2$ choose $2$ different flavors.... 
 A: If the four customers select three different flavors, then two of the four customers choose one of the $12$ flavors, a third customer chooses one of the remaining $11$ flavors, and the fourth customer chooses one of the remaining $10$ flavors, which can be done in  $$\binom{4}{2} \cdot P(12, 3) = 7920$$ ways.
As you noted, if the four customers select three of the four flavors, then either three of them choose one flavor while the fourth person chooses a different flavor or one pair chooses one flavor while another pair chooses a different flavor.
The number of ways three of the four customers can choose one of the $12$ flavors and the fourth chooses one of the remaining $11$ flavors is $$\binom{4}{3} \cdot P(12, 2) = 528$$ The number of ways one pair selects one of the $12$ available flavors while the other pair chooses one of the remaining $11$ flavors is $$\frac{1}{2} \cdot \binom{4}{2} \cdot P(12, 2) = 396$$  The factor of $1/2$ is necessary since the same selection results when the first pair chooses flavor $A$ then the second pair chooses flavor $B$ as when the second pair select flavor $B$ then the first pair chooses flavor $A$.  Thus, the number of ways that the customers can select two different flavors is $$\binom{4}{3} \cdot P(12, 2) + \frac{1}{2} \cdot \binom{4}{2} \cdot P(12, 2) = 528 + 396 = 924$$
Hence, the total number of ways of selecting either two or three different flavors is $$924 + 7920 = 8844$$ as you correctly inferred.  
