Probability: $n$ balls into $n$ holes with exactly one hole remaining empty The question is: n balls are distributed into n holes at random. What is the probability that exactly one hole remains empty.
I came up with $$P\left(A\right)=\frac{\:\dbinom{n\:}{1}\:\dbinom{n\:-1}{1}\:\left(n-2\right)!}{n^n}$$ but was told I’m way way off. Yes, I do realize I’m awful at probability.
Thanks for the help.
 A: Answer : $\frac{\binom{n}{2}n!}{n^n}$. You can reason about the answer this way : First put n balls in all n boxes (1 ball in each box). $n!$ ways to do that. Then, select any one ball ($n$ ways) and put in any of the other boxes($n-1$ ways). But here, every arrangement is being counted twice, so divide by 2. And to get the probability, divide by the total number of arrangements which is $n^n$.
A: There are indeed $n^n$ ways to distibute $n$ balls into $n$ holes: $n$ possible holes for each of the $n$ balls, independently of each other.
How many arrangements have exactly one hole empty: this means that there is one hole with two balls, and one empty hole. 
So pick the double hole (${n \choose 1} = n$ ways), pick the balls for that hole (${n \choose 2}$ ways), pick the empty hole $(n-1)$ ways, and we are left with $n-2$ balls to put in exactly $n-2$ holes (not the double one, not the empty one), and each must be filled, so we have $n-2$ choices for the first one, $n-3$ for the second, etc, so in total $(n-2)!$ ways to arrange the other balls.
So in total $\frac{n{n \choose 2}(n-1)(n-2)!}{n^n}$ as a probability.
A: $P\left(A\right)=\frac{\binom{n}{2}\binom{n}{1}\:\left(n-1\right)!}{n^n}$
For numerator, for $n$ distinct balls, choose two balls fall in the same hole, then choose the hole that is empty, then do the arrangement (treat the two balls which fall into the same hole as one element). 
For denominator, total possible arrangement for these $n$ balls with $n$ holes is $n^n$ (each ball has $n$ choices).
A: Other solutions appear to assume the balls are all unique.  I think it's more likely that the balls are indistinguishable and the holes are distinct.
There are $n(n-1)$ ways to put $n$ balls into $n$ holes with exactly one hole being empty.  There must be one hole with two balls.  So you have a choice of $n$ holes to be empty, then choose one of the $n-1$ remaining to put the extra ball in.
So how many ways are there to put $n$ balls into $n$ holes?  Imagine a binary string of $2n-1$ digits, where each $1$ represents a ball and each $0$ represents the boundary between holes.  Each binary string of length $2n-1$ containing $n$ $1$'s corresponds to a distinct distribution of $n$ balls among the $n$ holes.  There are obviously $\binom{2n-1}{n}$ such binary strings, so there are also $\binom{2n-1}{n}$ ways of distributing $n$ balls among $n$ holes.
So the final probability is:
$$\frac{n(n-1)}{\binom{2n-1}{n}} = \frac{n(n-1)}{\frac{(2n-1)!}{n!(n-1)!}} = \frac{n(n-1)n!(n-1)!}{(2n-1)!} = \frac{(n!)^2(n-1)}{(2n-1)!}$$
