Classic 10 Balls into 4 boxes problem Question:Throw at random 10 balls into 4 boxes. What is the probability that exactly 2 boxes remain empty?
My Solution: Using the stars and bars method, the boxes and balls can be thought as $0$ and | e.g. one configuration would be 
$$
 000000|00|0|0
$$
which would represent 6 balls in the first box etc. Thus the total number of configurations would be $\binom{13}{3}=286.$ 
There are $\binom{4}{2}$ to have 2 out of the 4 boxes empty, then to ensure the balls are put into only the other two boxes you place 1 ball into the other two boxes and then distribute the remaining 8 balls into the 2 boxes, which, using the same method as above, gives $\binom{9}{1}$ ways. Therefore there are $\binom{4}{2}*\binom{9}{1}=54$ ways of having exactly two boxes empty, giving a probability of $54/286\approx0.189.$ 
I did a simulation in MATLAB to confirm the answer, to find that I get a probability of $\approx0.005$, way off my answer. I can't see why my code and my solution above disagree so I'll post my code Here 
 A: Temporarily assume the balls are labelled.  This allows us to look at the sample space $\{1,2,3,4\}^{10}$ of size $4^{10}$ which will happen to be equiprobable, allowing us to use counting methods.  (the sample space of size $\binom{13}{3}=286$ where the balls are indistinct is not equiprobable and so we may not use elementary counting methods with this choice)
Break apart via multiplication principle.


*

*Pick which two boxes are empty.  $\binom{4}{2}$

*Pick how the balls are arranged among the two non-empty boxes.


*

*To do so, ignore the condition that they be non-empty: there are $2^{10}$ choices

*Then, remove the "bad" arrangements, where all ten are in the same box leaving the other empty.  There are $2$ bad arrangements

*This gives a total of $2^{10}-2$ ways to arrange among two boxes.



Applying multiplication principle then, there are $\binom{4}{2}(2^{10}-2)=6132$ ways to arrange the ten balls among the four boxes having exactly two boxes empty.
Dividing by the size of the sample space then, the probability is:
$$\frac{6132}{4^{10}}\approx 0.0058479$$
much more closely matching your simulation.
A: Generalizing (using urns instead of boxes, to make the variables a little easier to name):
How many ways are there to distribute $b$ unlabeled balls into $u$ urns such that exactly $k$ of those urns are empty?
Choose exactly $u-k$ urns that won’t be empty, group the $b$ balls into $u-k$ sets with at least $1$ ball in each group, and then count the number of ways to distribute those groups among the $u-k$ non-empty urns. 
To get the probability, divide by $u^b$, since each of the $b$ balls can be allocated among any of the $u$ urns.
$$P(b, u, k) = \frac{\binom{u}{u-k}{b \brace u-k}(u-k)!}{u^b}$$
Where ${n \brace k}$ is a Stirling number of the second kind.
$$P(10,4,2) = \frac{6132}{4^{10}} = 0.0058479309...$$
