Math Problem Statement

each student participates in exactly two lessons and any pair of students participates in exactly one lesson together. Show that there is a course in which at least 2/3 of the students participate.

Solution

To solve this, let’s use a combinatorial approach. We assume we have nn students, each participating in exactly two lessons, and each pair of students appears together in exactly one lesson. We want to show that there exists a course in which at least 23\frac{2}{3} of the students participate.

Step 1: Analyzing Participation and Pairs

  1. Let S={s1,s2,,sn}S = \{s_1, s_2, \dots, s_n\} be the set of students, and C={L1,L2,,Lk}C = \{L_1, L_2, \dots, L_k\} be the set of lessons.
  2. Since each student participates in exactly two lessons, each student appears in two distinct lessons from CC.
  3. For each pair of students, there is exactly one lesson they participate in together. This constraint implies a unique pairing structure across lessons, similar to the properties of a projective plane configuration or graph-theoretic design.

Step 2: Counting Total Lesson Slots

If each student is in exactly two lessons, the total "student-lesson slots" (or incidences) is 2n2n.

If each lesson LiL_i contains Li|L_i| students, we get: i=1kLi=2n.\sum_{i=1}^k |L_i| = 2n. Therefore, the total number of students across all lessons is twice the number of students in SS.

Step 3: Using the Pigeonhole Principle

Now, let's consider the average number of students per lesson. Since there are kk lessons and 2n2n student-lesson incidences, the average number of students per lesson is: 2nk.\frac{2n}{k}. To ensure that at least one lesson contains at least 23\frac{2}{3} of the students, we apply the pigeonhole principle.

Suppose each lesson contained fewer than 2n3\frac{2n}{3} students. Then: i=1kLi<k2n3=2n,\sum_{i=1}^k |L_i| < k \cdot \frac{2n}{3} = 2n, which contradicts the fact that the sum of all students in lessons is exactly 2n2n.

Conclusion

Thus, there must be at least one lesson containing at least 23\frac{2}{3} of the students, completing the proof.

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Related Questions

  1. What is the role of projective planes in combinatorial design problems like this one?
  2. How does the pigeonhole principle apply in combinatorial proofs?
  3. What are some real-world applications of combinatorial design principles?
  4. Could a similar principle be used if each student attended three lessons instead?
  5. How does this problem change if we relax the “exactly one lesson together” condition?

Tip

When tackling combinatorial proofs, break down the problem into manageable parts, like counting incidences or using averages, before applying general principles like pigeonhole

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Graph Theory
Projective Planes
Pigeonhole Principle

Formulas

Total student-lesson slots: 2n
Sum of students in all lessons: ∑ |L_i| = 2n
Average number of students per lesson: 2n / k

Theorems

Pigeonhole Principle
Combinatorial Design Theory

Suitable Grade Level

Undergraduate Level (Discrete Mathematics/Combinatorics)