Problem 231 - Combinatorics Through Guided Discovery

Question

A group of \(n\) students goes to a restaurant carrying backpacks. The manager invites everyone to check their backpack at the check desk and everyone does. While they are eating, a child playing in the check room randomly moves around the claim check stubs on the backpacks. We will try to compute the probability that, at the end of the meal, at least one student receives his or her own backpack. This probability is the fraction of the total number of ways to return the backpacks in which at least one student gets his or her own backpack back. (a) What is the total number of ways to pass back the backpacks? (b) In how many of the distributions of backpacks to students does at least one student get his or her own backpack? (c) What is the probability that at least one student gets the correct backpack? (d) What is the probability that no student gets his or her own backpack? (e) As the number of students becomes large, what does the probability that no student gets the correct backpack approach?

Step-by-Step Solution

Answer

The probability that at least one student gets the correct backpack is \( 1 - \frac{!n}{n!} \). The probability that no student gets the correct backpack approaches \( \frac{1}{e} \approx 0.3679 \) as \( n \) becomes very large.

Step 1: Total Number of Ways

The total number of ways to pass back the backpacks is given by the number of permutations of the backpacks, which is equal to the factorial of the number of students, denoted as \( n! \).

Step 2: Distributions with Correct Matching Backpack

To find the number of distributions where at least one student gets his or her own backpack, it is easier to first calculate the number of derangements (permutations where no student gets their own backpack). A derangement of \( n \) items is denoted as \( !n \). The number of distributions where at least one student gets his or her backpack is then calculated by subtracting the number of derangements from the total permutations: \( n! - !n \).

Step 3: Probability of Correct Backpack

The probability that at least one student gets the correct backpack is the fraction of favorable distributions over the total number of distributions. This can be expressed as: \[ P(\text{at least one correct}) = \frac{n! - !n}{n!} = 1 - \frac{!n}{n!} \]

Step 4: Probability of No Correct Backpack

The probability that no student gets his or her own backpack is simply the fraction of derangements over the total permutations, which can be written as: \[ P(\text{none correct}) = \frac{!n}{n!} \]

Step 5: Limit as Number of Students Increases

As the number of students \( n \) becomes very large, the number of derangements \( !n \) approaches \( \frac{n!}{e} \). Therefore, the probability that no student gets the correct backpack approaches: \[ P(\text{none correct}) \approx \frac{n!/e}{n!} = \frac{1}{e} \approx 0.3679 \]

Key Concepts

Combinatorial Probability

Combinatorial probability involves calculating the likelihood of certain events using combinatorial techniques. It often requires counting the number of ways events can happen and comparing it to the total possible outcomes.

This approach is handy when dealing with problems involving arrangements, combinations, or permutations. For example, in our exercise, we need to calculate the probability that at least one student gets their own backpack back.

We start by determining the total number of possible arrangements using permutations and then refine our calculations to focus on the desired event’s probability.

Permutations

Permutations are all the possible arrangements of a set of items where the order matters. For a set of items, the number of permutations is given by the factorial of n.In mathematical terms, if we have items, the number of permutations (denoted as n!) is:

!=1×2×3×...×(n−1)×n.

For our exercise, if we have students, the total number of ways to pass back the backpacks is the number of permutations of these backpacks or simply n!
,
:: permutations help us understand the possible arrangements and set the stage for calculating probabilities involving these permutations.

Factorial

A factorial, denoted as n!, is the product of all positive integers up to n.

It's a crucial concept in combinatorics. For example, 5!=1×2×3×4×5=120.

In our exercise, we use factorial to calculate the total number of permutations of the students' backpacks. If there are students, then there are (!ways to arrange the backpacks. This foundational calculation helps in further steps to determine probabilities related to specific arrangements.

Derangements

Derangements are permutations in which none of the items appear in their original positions. For example, if each student gets someone else's backpack, that's a derangement.

The number of derangements for n items is denoted as !$n.
Derangements can be calculated using the formula:
=!n=!n≅n!⎛1 −1!⎟(1+1i!,⎛⎛1+1⎛⎛+1!⎛⎛+⋯(−1)!.

In our exercise, we first calculate the number of derangements to find how many ways there are where no student gets their backpack. This lets us compute the probability that no one gets their original backpack.

Limit of Probability

As the number of students increases, the probability calculations approach a specific limit. Specifically, the probability that no one gets their correct backpack approaches:lim no right1e≅0..

This limit shows an intriguing result in combinatorial probability, suggesting that as grows large, the probability stabilizes around 3679.

Understanding this concept helps us predict outcomes in large sets and provides insight into the behavior of probabilities in combinatorial scenarios.