Find the probability of placing n balls into n numbered boxes so that exactly one box remains empty.
isn't the answer (nC2 * n!) / n^n

The problem you're referring to is about distributing \( n \) balls into \( n \) numbered boxes such that exactly one box remains empty.

### Step-by-Step Explanation:

1. **Total Possible Arrangements**: 
   Each of the \( n \) balls can be placed in any of the \( n \) boxes. So, the total number of ways to place \( n \) balls into \( n \) boxes is \( n^n \).

2. **Arrangements with Exactly One Empty Box**:
   If exactly one box remains empty, then the other \( n-1 \) boxes must contain all \( n \) balls. This is essentially a surjection problem, where we need to distribute \( n \) balls into \( n-1 \) boxes, ensuring that each box gets at least one ball.

   To compute this, the steps are as follows:
   - Choose which box will remain empty. There are \( \binom{n}{1} = n \) ways to choose the empty box.
   - Now, distribute \( n \) balls into the remaining \( n-1 \) boxes, such that no box is empty. This is a classical "surjection" problem, and the number of ways to do this is given by **Stirling numbers of the second kind** multiplied by factorial of \( n-1 \). This gives the count of surjective functions.

### So your initial assumption:
   \[
   \frac{\binom{n}{2} \cdot n!}{n^n}
   \]
   is incorrect because you are not taking the surjection into account properly.

To correctly solve this, we need to involve Stirling numbers. Would you like to dive deeper into that aspect, or clarify the role of Stirling numbers?

Find the probability of placing n balls into n numbered boxes so that exactly one box remains empty. Isn't the answer (nC2 * n!) / n^n?

Learn how to solve the probability problem of distributing n balls into n numbered boxes so that exactly one box remains empty. This problem involves combinatorics, surjection, and Stirling numbers of the second kind. Suitable for advanced high school or undergraduate students.

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