Find the probability of placing n balls into n numbered boxes so that exactly
one box remains empty.


The problem is to find the probability of placing \( n \) distinguishable balls into \( n \) distinguishable numbered boxes such that exactly one box remains empty. 

### Steps to solve:

1. **Total number of possible arrangements**:  
   Each of the \( n \) balls can be placed into one of the \( n \) boxes. Since there are no restrictions initially, the total number of possible ways to assign \( n \) balls into \( n \) boxes is \( n^n \).

2. **Arrangements with exactly one empty box**:
   - To have exactly one empty box, select one box to remain empty. The number of ways to choose which box remains empty is \( \binom{n}{1} = n \).
   - Now, we need to place \( n \) balls into the remaining \( n - 1 \) boxes, such that none of these boxes is empty. This is a **surjection problem** or an "onto" problem, where each of the \( n-1 \) boxes must have at least one ball.
   - The number of ways to distribute \( n \) balls into \( n-1 \) boxes with none left empty is given by the **Stirling number of the second kind** multiplied by \( (n-1)! \), which accounts for assigning the balls into the boxes. The number of such distributions is \( S(n, n-1) \times (n-1)! \), where \( S(n, n-1) \) is the Stirling number of the second kind.

   The Stirling number \( S(n, n-1) = \binom{n}{2} = \frac{n(n-1)}{2} \). So the number of favorable ways to distribute the balls is \( \frac{n(n-1)}{2} \times (n-1)! \).

3. **Probability**:
   The probability is the ratio of favorable outcomes to total outcomes:

   \[
   P(\text{exactly one empty box}) = \frac{n \times S(n, n-1) \times (n-1)!}{n^n}
   \]

   Substituting \( S(n, n-1) = \binom{n}{2} \):

   \[
   P = \frac{n \times \frac{n(n-1)}{2} \times (n-1)!}{n^n}
   \]

   Simplifying the expression:

   \[
   P = \frac{n \times \frac{n(n-1)}{2} \times (n-1)!}{n^n}
   \]

   Further simplification gives the final probability.

---

Would you like to see the expanded details of the Stirling numbers or any further clarifications?

Here are 5 related questions you might explore:

1. How do Stirling numbers of the second kind relate to other combinatorial problems?
2. What is the probability that no boxes are left empty when placing \( n \) balls into \( n \) boxes?
3. How does the probability change as \( n \) increases?
4. Can you calculate the expected number of empty boxes when placing \( n \) balls into \( n \) boxes?
5. How does the problem change if the boxes are indistinguishable?

**Tip**: Stirling numbers of the second kind are very useful in partition problems, especially for distributing objects into groups with specific conditions!

Find the probability of placing n balls into n numbered boxes so that exactly one box remains empty.

Learn how to find the probability of placing n balls into n numbered boxes with exactly one box remaining empty. This problem explores combinatorics, probability, and the use of Stirling numbers of the second kind, suitable for undergraduate-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation