Doubt about Probability of arranging identical balls There are four boxes and 12 balls. The boxes are numbered and hence distinguishable but the balls are identical. 
What is the probability that a random arrangement would result in 10 balls in box 1 2 ball in box 2 and the rest boxes are empty ?
My attempt was solving $x_1+x_2+x_3+x_4 = 12$ and only one case is favourable.
My friend's attempt - assume the balls are numbered, total number of arrangements is $4^{12}$ out of which $\frac{12!}{10!2!}$ are favourable.
Whose method is correct?
EDIT
Okay if we choose to throw the balls then the probability comes out as $$\frac{\frac{12!}{10!2!}}{4^{12}}$$.
Now if the balls are unidentical then also the answer is $\frac{\frac{12!}{10!2!}}{4^{12}}$. How is it happening ? 
 A: Consider having the 12 identical balls, and three identical "dividers" arranged in a straight line. (Up to the first divider is Box #1, between the first and second dividers  is Box #2, and so on. No need for an end-of-Box #4 marker)
How many ways can you arrange these 15 items?
How many of them have the first divider after the tenth ball, and the rest after the twelfth ball?
A: As I see it, and correct me if I misunderstood the question, the number of balls in each box is absolutely random. This means that each box has the same probability of having any number of balls between $0$ and $12$. With this in mind we now have to take into account that the total number of balls we have is $12$. Your approach is correct, but brute forcing it by finding all positive integer solutions that satisfy the equation is hard to do by hand. (You could write a computer program instead. I'll do it when I have a moment and edit the answer.)
However, I think we can break the problem in two parts:
First we'll try to get the probability of getting $10$ balls in box one and then add it to the probality of getting $2$ balls in box $2$.
If the arrangement is random, in the first box we could get any number of balls ranging from $0$ to $12$, so the possibility of getting $10$ balls is:
$P(X_1=10)=\frac{1}{13}$
After that, getting them in box 1 means that $P(X=10)$ gets reduced by $4$ as just $1$ in $4$ cases is favourable.
So:
$P(X_1=10)=\frac{1}{13·4}$
Now we have to add the probability of getting two balls in box $2$. We will assume that we got $10$ balls on box $1$, so the probability then is:
$P(Y_2=2)=\frac{1}{3}·\frac{1}{3}$
Adding the two probabilities together we get that the total probability is:
$P(X_1 and Y_2)=\frac{1}{13·4}·\frac{1}{9}=\frac{1}{468}=0.002137$
Edit:
I brute forced the answer with a Python program which resolved the problem in the way you suggested. The code is as follows:
def prob():
 results=[] #stored valid solutions
 Total=0 #total of iterarions
 Valid=0 #count valid solutions
 for i in range(13): #ranging from 0 to 12 balls in box 1
     for j in range(13): #ranging from 0 to 12 balls in box 2
         for k in range(13): #ranging from 0 to 12 balls in box 3
             for l in range(13): #ranging from 0 to 12 balls in box 4
                 if i+j+k+l==12: #check if it is a valid solution
                    results.append((i,j,k,l)) #if it is, append to solutions
                    Total+=1;Valid+=1 #keep count of valid solutions
                 else:
                    Total+=1 #total iterations
 return results,Total,Valid

The program should run $13⁴=28561$ iterations (which does) and check how many of them are valid solutions. Once we get this number, as there is only one valid solution the probability must be $P(x_1andY_2)=\frac{1}{\text{solutions}}$. 

Each $(i,j,k,l)$ term stands for a valid solution and the two numbers at the end are total iterations and valid solutions.
The program outputs $455$ solutions, hence the probability we are looking for is:
$P(X_1andY_2)=\frac{1}{455}=0.002198$ which is a bit higher (13 cases up) than the one I previously calculated.The point is I don't know what I am missing.
A: There can be disagreement about the appropriate model. The model I would favour has us "throwing" the balls one at a time towards the boxes, with a ball equally likely to fall in any of the boxes, and the results of the $12$ throws independent. It is useful to imagine that the balls have ID numbers $1$ to $12$ on them, if you prefer in invisible ink. These ID numbers do not affect the probabilities.
So the appropriate sample space under this model has $4^{12}$ equally likely outcomes. 
Now we count the "favourables." Call the boxes A, B, C, D. The favourables are the words of length $12$ that have $10$ A's and $2$ B's. The location of the $10$ A's can be chosen in $\binom{12}{10}$ ways, and now the location of the B's is determined. 
Your friend is clearly using the same "throwing" model as the one used in this answer. 
Note that the answer we get is quite different from the one given by an analysis that makes all solutions of $x_1+x_2+x_3+x_4=12$ in non-negative integers equally likely. That model seems quite unsuitable, for example, in the standard application of this sort of balls and boxes problem to hashing. 
