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Each of balls is independently placed into one of boxes, with all boxes equally likely. What is the probability that exactly one box is empty?

Yes, I am well aware that the solution has been posted. While they make sense to me, I have an incorrect solution that I can't seem to point out how it's wrong.

Here is my logic for the numerator:

  • We have $n$ ways to choose an empty bin
  • Exactly 1 ball will contribute to an already non-empty box (making that box have 2 balls instead of 1). There are $n$ ways to choose the ball, and $n - 1$ ways to choose the box.
  • For the remaining $n - 1$ balls, there are $(n - 1)!$ ways to permute them across the $n - 1$ boxes, to guarantee that all the boxes are nonempty.

The numerator simplifies to $n(n - 1)n!$. Comparing with the correct solution of $\binom{n}{2} \cdot n!$, I am overcounting by a factor of 2. Could someone explain to me how?

Umi Sumi
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    Think about the two balls in the single urn. You have no way to decide which is those is the "extra" one you consider in step two so you count that given pattern twice. – lulu Aug 30 '22 at 07:36
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    Note: a good way to trap errors like this is to write it all out for some small $n$. Taking $n=2$ is good enough here. – lulu Aug 30 '22 at 07:42
  • Thanks @lulu, your explanation makes sense :) – Umi Sumi Aug 30 '22 at 17:55

1 Answers1

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Assume that the result is the No.1 ball and No.2 ball is in the same box.Then when you choose No.1 ball for the lucky ball, you calculate this situation once.Then conversely,when you choose No.2 ball for the lucky ball,for the exact same situation,you calculate it for another time.So your answer is two times of the right one.

Re-ocean
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  • Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Community Aug 30 '22 at 07:41