What is the correct way to think about this yet another balls/boxes problem? How would you do the following problem:

Suppose that $n$ balls are placed at random into $n$ boxes. Find the probability that there is exactly one empty box.

I mentally pictured $n$ boxes being in front of me and $n$ balls which I'm throwing (hence randomness) in the boxes, each ball has an equal chance of landing in any of the boxes. That led me to this: $$\frac{n(n-1)\frac{n!}{2}}{n^n}\tag{1}$$
Which is a good answer yet to a different problem, and the answer to the above problem turned out to be
$$\frac{n(n-1)}{\binom{2n-1}{n}}\tag{2}$$
I can reverse engineer the correct answer $(2)$ and think about it in terms of stars and bars but it's unclear what mental picture do I need to have in mind after reading the word problem?
And as a side question: do you think the phrasing of the cited problem is unambiguous? How would you state the problem which has $(1)$ as its solution?
 A: There is some ambiguity about the statement of the problem, since "randomly" can be interpreted in many ways. 
We interpret the statement as meaning that the balls are placed in boxes one at a time. Each time, a ball is equally likely to end up in any of the boxes, and the choices are independent.
To visualize, it is useful to consider the balls to be distinct. That does not affect the probability. 
For each ball, we have $n$ choices for the box it goes into. That gives a total of $n^n$ equally likely choices.
Now we count the number of ways to have precisely one empty box. That unlucky box can be chosen in $n$ ways. There will then be a lucky box that gets $2$ balls. That lucky box can be chosen in $n-1$ ways.
The balls that end up in the lucky box can be chosen in $\binom{n}{2}$ ways. For each of these ways, the $n-2$ boxes that will receive exactly $1$ ball can be filled from the $n-2$ remaining balls in $(n-2)!$ ways. 
That gives a total of $n(n-1)\binom{n}{2}(n-2)!$ "favourables." This simplifies to $\binom{n}{2}n!$. Divide by $n^n$ for the probability. 
Remark: By "Stars and Bars" there are $\binom{2n-1}{n-1}$ of distributing $n$ identical balls among $n$ boxes. Let us use probability model that says all these ways are equally likely.
Now we count the favourables. The unlucky box can be chosen in $n$ ways, and for each way the lucky box can be chosen in $n-1$ ways. Now the distribution is determined, so there are $n(n-1)$ favourables. 
The model we used for the second calculation is physically very unreasonable. It is very different from the first model. Under the first model, all balls ending up in the first box has probability $\frac{1}{n^n}$. Under the second, the probability is $\frac{1}{\binom{2n-1}{n-1}}$, a much larger number for large $n$.
The second model does not seem suitable for any application I can think of. (The first is important in many places, including the study of hashing.)
I am of the opinion that the first model is the most reasonable interpretation of the problem. 
A: Number the boxes $B_1$, $B_2,\,\dots B_n$, and the balls $b_1$, $b_2,\,\dots b_n$.
Then the probability that there is exactly one empty box is the number of arrangements in which there is exactly one empty box, divided by the total number of possible arrangements. 

The number of arrangements in which there is exactly one empty box is the number of ways to put $n$ balls in $n-1$ boxes, multiplied by $n$ (the number of ways to choose which box is empty).

We want to put $n$ balls in $n-1$ boxes, such that there are no empty boxes. There are $n-1$ ways to choose the box in which there are two balls. Then there are $\binom{n}{2}$ ways to choose the two balls that go in this box. Finally, there are $(n-2)!$ ways to distribute the remaining $n-2$ balls in the remaining $n-2$ boxes. Multiplying these terms together, we get:
$$(n-1)\binom{n}{2}(n-2)! = \frac{n! \cdot n \cdot (n-1)}{2}$$

The total number of arrangements is $n^n$ (each ball has $n$ possible boxes into which it can be placed). Thus, our final probability is:
$$\frac{n! \cdot n \cdot (n-1)}{2(n^n)}$$
A: Here is another way of thinking with multinomial coefficient. The answer would be 
$$ \frac{\displaystyle {n \choose 0~ 2~  1~ \cdots 1} \frac{1}{(n-2)!} \, n!}{\displaystyle n^n}.$$
For the denominator, consider each ball, which has $n$ possible choices of boxes to go. There are $n$ balls, which yields $n \cdot n \cdots n = n^n$ ways in total.  
For the numerator, first note that the $n$ boxes must have $0, 2, 1, \ldots, 1$ balls, respectively, since there is only one empty ball. We may first divide $n$ balls into $n$ groups so that one group contains 0, next contains 2, all the remaining $(n-2)$ groups each have only 1 ball. According to the meaning of multinomial coefficient, there are ${n \choose 0~ 2~ 1~ \cdots 1}$ ways of doing this if these groups are all distinct. However, those $(n-2)$ groups (those with one ball) are not distinct, explaining the readjustment factor $1/(n-2)!$. Next there are $n!$ ways of assigning these $n$ groups to $n$ boxes. Therefore, the result is further multiplied by $n!.$ 
After simplification, you should get the same final answer $n(n-1) n!/(2 n^n).$
A: I think there are two lines of reasoning here: 


*

*dealing with undistinguishable balls

*dealing with distinguishable balls


Let's take the first scenario, where the balls are undistinguishable.
The number of ways you can distribute n balls in n boxes is given by the Bose-Einstein formula. In this scenario you don't care which balls land in which boxes. In order to help with intuition think of not having a way to make a difference between the balls. 
This gives: 
$$\binom{n + n - 1}{n - 1}$$
In this case, you don't care which balls will end up together in a box, you only care which box is empty and which box has 2 balls. For that, there are $\binom{n}{1}$ ways to pick an empty box and from the remaining $n-1$ boxes there are $\binom{n-1}{1}$ ways to pick the box that will have the two balls in it. The rest, you can assume it will be one ball per box. 
All of this can be translated into: 
$$\frac{\binom{n}{1}\binom{n-1}{1}}{\binom{2n-1}{n-1}}$$
In case the balls are distinguishable, it's the same as having a number on every ball. Here, you care which balls go in which box and every combination makes another case that you need to count. If the first and 3rd ball goes in the first box, is not the same as first and second going in the first box, even if you're using the same box. 
There are $n^n$ ways to put n distinguishable balls in n boxes. (For intuition: First ball can go in any of the n boxes, second can go in any of the n boxes, and so on). 
The way you pick the empty box and the box that has two balls is the same, however it matters which balls go in the box that contains 2 balls, and also it matters how you put the balls in the remaining $n-2$ boxes. 
Picking two balls out of n is given by $\binom{n}{2}$ and arranging the rest of $n-2$ balls in $n-2$ boxes can be done in $(n-2)!$ way. (As intuition, think that you have your boxes numbered from $1$ to $n-2$, you can permute the balls in $(n-2)!$ ways and put the ball in boxes in that particular order).
If everything is being put together this gives: 
$$\frac{\binom{n}{1}\binom{n-1}{1}\binom{n}{2}(n-2)!}{n^n}$$
I think the statement is not mentioning if you're dealing with distinguishable/non distinguishable balls and I'd assume that it's being expected to provide the reasoning around both cases. 
