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

We are tasked with finding the probability of placing \(n\) balls into \(n\) numbered boxes such that exactly one box remains empty.

### Step 1: Total Number of Possible Arrangements
If we have \(n\) balls and \(n\) numbered boxes, the total number of ways to distribute the \(n\) balls into the \(n\) boxes (with no restrictions) is:

\[
n^n
\]

This is because for each ball, there are \(n\) choices (one for each box), and there are \(n\) balls in total.

### Step 2: Number of Favorable Outcomes (Exactly One Box Empty)
We now calculate the number of favorable outcomes, where exactly one box is empty.

1. **Choose which box will be empty**: There are \(n\) boxes, so there are \(n\) ways to choose which box remains empty.

2. **Distribute \(n\) balls into the remaining \(n-1\) boxes**: Once one box is empty, we need to distribute all \(n\) balls into the remaining \(n-1\) boxes, with the condition that none of these boxes is empty.

The number of ways to distribute \(n\) balls into \(n-1\) non-empty boxes is the same as the number of **surjections** (onto functions) from \(n\) balls to \(n-1\) boxes. The number of surjections from a set of size \(n\) to a set of size \(n-1\) is given by:

\[
S(n, n-1) = (n-1)! \cdot \left( \binom{n}{1} - \binom{n}{2} + \binom{n}{3} - \cdots + (-1)^{n-1} \binom{n}{n-1} \right)
\]

This is the inclusion-exclusion formula for surjections.

### Step 3: Probability Calculation
The probability is the ratio of the number of favorable outcomes to the total number of outcomes:

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

### Step 4: Simplifying with Stirling Numbers
The term \(S(n, n-1)\) is known to simplify to:

\[
S(n, n-1) = (n-1)^{n-1}
\]

Thus, the probability becomes:

\[
P(\text{one box empty}) = \frac{n \cdot (n-1)^{n-1}}{n^n} = \frac{(n-1)^{n-1}}{n^{n-1}}
\]

### Final Answer:
The probability of placing \(n\) balls into \(n\) numbered boxes such that exactly one box remains empty is:

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

Would you like more details on this explanation or any clarifications? 

Here are 5 related questions:
1. What is the inclusion-exclusion principle used in counting?
2. How are surjections related to Stirling numbers of the second kind?
3. Can this probability result be extended to cases where more than one box is empty?
4. How would the result change if the balls were indistinguishable?
5. What is the significance of Stirling numbers in combinatorics?

**Tip**: The inclusion-exclusion principle is a powerful tool in combinatorics for avoiding overcounting by accounting for intersections of sets.

Learn how to find the probability of placing n balls into n numbered boxes such that exactly one box remains empty. This problem involves combinatorics, the inclusion-exclusion principle, surjections, and Stirling numbers. Suitable for advanced high school or undergraduate 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