No box contains the same two letters

Oct 13, 2025 by talynt.org 20 views

Introduction

In combinatorics, the study of counting and arranging objects in various ways is a fundamental concept. This article explores a specific problem involving the distribution of identical letters in different boxes, ensuring that no box contains the same two letters. We will delve into the details of this problem, analyze the constraints, and provide a solution using combinatorial techniques.

Problem Statement

We have 8 letters: $a,a,b,b,c,c,d,d$ , where the two $a$ 's are identical, and so are the other pairs of letters. We have 4 different boxes, denoted as $x_1,x_2,x_3,x_4$ . The goal is to distribute the 8 letters in the 4 boxes such that no box contains the same two letters.

Constraints and Analysis

The problem has two main constraints:

No box contains the same two letters: This means that each box must have a unique pair of letters.
4 different boxes: The boxes are distinct, and we cannot have multiple boxes with the same label.

To analyze the problem, let's consider the possible arrangements of letters in each box. Since the letters are identical, we can treat the pairs of letters as single entities. This simplifies the problem and allows us to focus on the arrangement of these entities in the boxes.

Combinatorial Approach

One way to approach this problem is to use the concept of Stirling numbers of the second kind. These numbers count the number of ways to partition a set of distinct objects into non-empty subsets. In this case, we want to partition the 8 letters into 4 non-empty subsets, where each subset represents a box.

The Stirling number of the second kind, denoted as $S(n,k)$ , counts the number of ways to partition a set of $n$ distinct objects into $k$ non-empty subsets. In our case, we have $n=8$ and $k=4$ . The Stirling number of the second kind can be calculated using the following formula:

S(n,k) = \frac{1}{k!} \sum_{i=0}^{k} (-1)^i \binom{k}{i} (k-i)^n

Using this formula, we can calculate the number of ways to partition the 8 letters into 4 non-empty subsets.

Calculating the Number of Ways

To calculate the number of ways to partition the 8 letters into 4 non-empty subsets, we need to evaluate the Stirling number of the second kind for $n=8$ and $k=4$ . Plugging in the values, we get:

S(8,4) = \frac{1}{4!} \sum_{i=0}^{4} (-1)^i \binom{4}{i} (4-i)^8

Evaluating the sum, we get:

S(8,4) = 1701

This means that there are 1701 ways to partition the 8 letters into 4 non-empty subsets.

Arranging the Letters in the Boxes

Now that we have the number of ways to partition the letters into 4 non-empty subsets, we need to arrange the letters in the boxes. Since the letters are identical, we can treat the pairs of letters as single entities. This simplifies the problem and allows us to focus on the arrangement of these entities in the boxes.

To arrange the letters in the boxes, we can use the concept of permutations. A permutation is an arrangement of objects in a specific order. In this case, we want to arrange the 4 entities (pairs of letters) in the 4 boxes.

The number of permutations of $n$ objects is given by $n!$ . In our case, we have $n=4$ , so the number of permutations is:

4! = 24

This means that there are 24 ways to arrange the 4 entities in the 4 boxes.

Combining the Results

To find the total number of ways to distribute the 8 letters in the 4 boxes, we need to combine the results from the previous steps. We have 1701 ways to partition the letters into 4 non-empty subsets, and 24 ways to arrange the letters in the boxes.

Since the partitioning and arrangement are independent events, we can multiply the number of ways to get the total number of ways:

1701 \times 24 = 40824

This means that there are 40824 ways to distribute the 8 letters in the 4 boxes, such that no box contains the same two letters.

Conclusion

In this article, we explored a combinatorial problem involving the distribution of identical letters in different boxes. We analyzed the constraints, used the concept of Stirling numbers of the second kind to partition the letters into non-empty subsets, and arranged the letters in the boxes using permutations. By combining the results, we found that there are 40824 ways to distribute the 8 letters in the 4 boxes, such that no box contains the same two letters.

Q: What is the main goal of the problem?

A: The main goal of the problem is to distribute the 8 letters ( $a,a,b,b,c,c,d,d$ ) in 4 different boxes ( $x_1,x_2,x_3,x_4$ ) such that no box contains the same two letters.

Q: What are the constraints of the problem?

A: The two main constraints of the problem are:

No box contains the same two letters: Each box must have a unique pair of letters.
4 different boxes: The boxes are distinct, and we cannot have multiple boxes with the same label.

Q: How do we approach this problem?

A: We use the concept of Stirling numbers of the second kind to partition the 8 letters into 4 non-empty subsets, where each subset represents a box. We then arrange the letters in the boxes using permutations.

Q: What is the significance of Stirling numbers of the second kind?

A: Stirling numbers of the second kind count the number of ways to partition a set of distinct objects into non-empty subsets. In this case, we use them to partition the 8 letters into 4 non-empty subsets.

Q: How do we calculate the number of ways to partition the letters?

A: We use the formula for Stirling numbers of the second kind:

S(n,k) = \frac{1}{k!} \sum_{i=0}^{k} (-1)^i \binom{k}{i} (k-i)^n

where $n=8$ and $k=4$ .

Q: What is the result of calculating the number of ways to partition the letters?

A: The result is $S(8,4) = 1701$ , which means there are 1701 ways to partition the 8 letters into 4 non-empty subsets.

Q: How do we arrange the letters in the boxes?

A: We use permutations to arrange the 4 entities (pairs of letters) in the 4 boxes.

Q: What is the significance of permutations in this problem?

A: Permutations allow us to arrange the letters in the boxes in a specific order.

Q: How do we calculate the number of ways to arrange the letters in the boxes?

A: We use the formula for permutations:

n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1

where $n=4$ .

Q: What is the result of calculating the number of ways to arrange the letters in the boxes?

A: The result is $4! = 24$ , which means there are 24 ways to arrange the letters in the boxes.

Q: How do we combine the results to find the total number of ways to distribute the letters?

A: We multiply the number of ways to partition the letters (1701) by the number of ways to arrange the letters in the boxes (24) to get the total number of ways:

1701 \times 24 = 40824

Q: What is the final answer to the problem?

A: The final answer is 40824, which is the total number of ways to distribute the 8 letters in the 4 boxes such that no box contains the same two letters.