Understanding Student Participation in Competitions

This problem involves analyzing student participation in three different competitions: Quiz, Extempore, and Debate. We are given information about the number of students participating in various combinations of these competitions and need to find the number of students who participated only in the Debate competition.

Breaking Down the Student Participation Data

Let's denote the sets of students participating in Quiz, Extempore, and Debate as Q, E, and D, respectively. The total number of students is 60, and each student participated in at least one competition. This means the total number of students is the size of the union of the three sets: $|Q \cup E \cup D| = 60$.

The total number of students participating in one or more competitions can be divided into categories based on the exact number of competitions they participated in:

• Students who participated in only one competition (Quiz only, Extempore only, or Debate only).
• Students who participated in exactly two competitions (Quiz and Extempore only, Extempore and Debate only, or Quiz and Debate only).
• Students who participated in all three competitions (Quiz, Extempore, and Debate).

The sum of students in these mutually exclusive categories equals the total number of students who participated in one or more competitions.

Analyzing the Given Information

We are given the following specific values:

• Total students $|Q \cup E \cup D| = 60$.
• Students participated either in Quiz only or in Extempore only: $|Q \text{ only}| + |E \text{ only}| = 22$.
• Students participated in all three competitions: $|Q \cap E \cap D| = 4$.
• Students participated in any of the two competitions only: $|(Q \cap E) \text{ only}| + |(E \cap D) \text{ only}| + |(Q \cap D) \text{ only}| = 14$.

We want to find the number of students who participated in Debate only, which is $|D \text{ only}|$.

Calculating Students in Debate Only

Using the principle that the total number of students is the sum of students in each distinct region of the Venn diagram, we can write the equation:

Substitute the known values into this equation:

Now, let's simplify the equation by adding the known numbers on the right side:

To find the number of students who participated in Debate only, subtract 40 from 60:

Therefore, 20 students participated only in the Debate competition.

Conclusion on Competition Participation

Based on the provided data and calculation, the number of students who participated in Debate only is 20.

Revision Table: Competition Participation Analysis

Additional Information: Venn Diagrams and Set Theory Basics

This problem is a classic example that can be solved using the principles of set theory, specifically Venn diagrams. A Venn diagram visually represents sets and their relationships, including intersections and unions.

• Union ($A \cup B \cup C$): Represents the total number of elements in sets A, B, or C, covering all regions within the circles. In this problem, it's the total number of students who participated in at least one competition.
• Intersection ($A \cap B \cap C$): Represents the elements common to all sets A, B, and C. In this problem, it's the students who participated in all three competitions.
• Region "Only": Refers to elements that belong exclusively to one set or the intersection of a specific combination of sets, excluding elements from other sets. For example, "Debate only" means students in Debate but not in Quiz or Extempore. "Quiz and Extempore only" means students in both Quiz and Extempore but not in Debate.

The formula used to solve this problem leverages the idea that the total in the union is the sum of the counts in each distinct region (only one, exactly two, exactly three). This approach avoids the complexity of the full Principle of Inclusion-Exclusion formula for three sets ($|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |A \cap C|) + |A \cap B \cap C|$) by working directly with the counts of the specific regions defined in the problem.