Probability - balls and rooms

Question

We have to enter $n$ different balls in $n$ different rooms randomly. Which is the probability to left exactly one room empty?

If it's possible I have to solve this problem using the probability formulas and theorems, not with combinatory formulas.

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I don't know if I'm complicating the problem but what I have done is the following: I know that lefting exactly one room empty is equal to entering 2 balls in a room and one ball on the other $n-2$ rooms. So firstly, I've calculated the number of ways to choose the 2 balls which I would enter in the same room; for that, we've got $\frac{n!}{2!(n-2)!}$ ways. Then, I've tried to sort the other $n-2$ balls, the 2 balls which are toghether and the empty box; so as that is the same as sorting $n$ elements, we've got $n!$ ways for it. So I finally get that the result is $\frac{(n!)^2}{2!(n-2)!}$

Anyway, I'm not sure the result I obtain is the correct one... Moreover, I haven't use any probability formula or theorem...

Any idea of how can I solve it?

Answer 1

There are $n\choose2$ ways to choose the two balls which go in the same room, $n$ ways to choose the room with two balls and $n-1$ ways to choose the empty room after the room with two balls has been chosen. There are $(n-2)!$ ways to assign the remaining $n-2$ balls among the remaining $n-2$ rooms.

So there are $n(n-1){n\choose 2}(n-2)!=n^2(n-1)^2(n-2)!/2$ ways to assign the balls so that exactly one room is left empty. The total number of ways to assign the balls to rooms is $n^n$.

The desired probability is $\frac{(n-1)^2(n-2)!}{2n^{n-2}}$

Answer 2

There are $n^n$ equiprobable ways to distribute the $n$ balls randomly in the $n$ rooms.

We now have to count the admissible distributions. There are $n!$ ways where each room obtains exactly one ball. Given such a perfect distribution you can select an arbitrary room in $n$ ways and then put its ball into one of the other $n-1$ rooms, in order to obtain an admissible distribution. Since exactly $2$ perfect distributions will give rise to the same admissible distribution we see that there are ${1\over2}n!\> n(n-1)$ admissible distributions. It follows that your probability is $$p={n!\>n(n-1)\over 2\>n^n}\ .$$