Hello Members,
If n balls are placed at random into n cells, How to find the probability that exact one cell remains empty?
Answer:-
n balls can be placed into n cells in $n^n$ ways.How to proceed further?
Are both the balls and the cells distinguishable? If they are, then you are correct that there are $n^n$ ways to place the $n$ balls into $n$ cells. Out of the $n$ cells, choose one to be empty. Of the remaining $n-1$ cells, choose one of them to have two balls. Then the probability is $\dfrac{n(n-1)}{n^n}$. Chances are, the balls are not distinguishable but the cells are. In that case, there are $\dbinom{2n-1}{n}$ ways to place $n$ indistinguishable balls into $n$ distinguishable cells. There are $\dbinom{n-1}{1}$ ways to place $n$ indistinguishable balls into $n-1$ distinguishable cells so that every cell has at least one ball. Therefore, the probability is $\dfrac{n-1}{\dbinom{2n-1}{n}}$. There are also the possibilities that the balls are distinguishable but the cells are not and that the neither the balls nor the cells are distinguishable.
My answer was incorrect. Let's try again. Suppose $n=1$. Then there is a zero percent chance that exactly one cell will be empty and a 100% chance that every cell will have one ball (one ball, one cell, there is only one way to place the balls). So, assume $n>1$. As I mentioned above, we choose an empty cell and we choose a cell to have two balls. There are $n(n-1)$ ways of choosing that. Then, we choose two of the balls to be the two balls in the cell that has two balls. There are $\dbinom{n}{2}$ ways of choosing that. This leaves $n-2$ cells and $n-2$ balls. So, basically, we order the remaining $n-2$ balls and place them into the remaining $n-2$ cells in order. That is $(n-2)!$. Since each step was independent of the previous, we apply the product rule. Thus, we have:
$n(n-1)\dbinom{n}{2}(n-2)!$ ways to place $n$ distinguishable balls in $n$ distinguishable cells with exactly one cell remaining empty. That is equal to $n!\dbinom{n}{2}$, just as the book has. My mistake. I went too quickly through the problem. Sorry about that.