In a class of 45 students, all students participate in one or both of the two games, i.e. Chess and Badminton on. 11 students participate in both the games, whereas 17 students participate only in Chess. How many total students participate in Badminton?
28
Understanding the Student Participation Problem
This problem involves determining the number of students who participate in a specific game, Badminton, within a group where all students participate in at least one of two games: Chess or Badminton. We are given information about the total number of students, those who play only Chess, and those who play both games.
Given Information
- Total number of students in the class = 45
- All students participate in either Chess or Badminton or both.
- Number of students who participate in both Chess and Badminton = 11
- Number of students who participate only in Chess = 17
Goal
Find the total number of students who participate in Badminton. This includes students who play only Badminton and students who play both Chess and Badminton.
Applying Set Theory Concepts
We can think of this problem using basic set theory, even without drawing a full Venn diagram. The total number of students is the sum of those who play only Chess, those who play only Badminton, and those who play both.
The relationship can be expressed as:
$\text{Total Students} = (\text{Students only in Chess}) + (\text{Students only in Badminton}) + (\text{Students in both})$
Step-by-Step Calculation
Step 1: Find the number of students who participate only in Badminton.
We know the total students, students only in Chess, and students in both. We can rearrange the formula to find the students only in Badminton:
$\text{Students only in Badminton} = \text{Total Students} - (\text{Students only in Chess}) - (\text{Students in both})$
Plugging in the given values:
$\text{Students only in Badminton} = 45 - 17 - 11$
$\text{Students only in Badminton} = 45 - 28$
$\text{Students only in Badminton} = 17$
So, 17 students participate only in Badminton.
Step 2: Find the total number of students who participate in Badminton.
The total number of students who participate in Badminton includes those who play only Badminton and those who play both Chess and Badminton.
$\text{Total students in Badminton} = (\text{Students only in Badminton}) + (\text{Students in both})$
Using the value calculated in Step 1 and the given value for students in both:
$\text{Total students in Badminton} = 17 + 11$
$\text{Total students in Badminton} = 28$
Therefore, a total of 28 students participate in Badminton.
Summary of Student Participation
| Category | Number of Students |
|---|---|
| Only Chess | 17 |
| Only Badminton | 17 |
| Both Chess and Badminton | 11 |
| Total Students | $\textbf{17 + 17 + 11 = 45}$ |
| Total Badminton Participants | $\textbf{17 (Only Badminton) + 11 (Both) = 28}$ |
| Total Chess Participants | $\textbf{17 (Only Chess) + 11 (Both) = 28}$ |
The total number of students participating in Badminton is 28.
Revision Table: Key Concepts in Set Theory Problems
| Concept | Explanation | Formula (for two sets A and B) |
|---|---|---|
| Union ($A \cup B$) | Elements in set A, or set B, or both. In this problem, this is the total number of students (45), as everyone plays at least one game. | $|A \cup B| = |A| + |B| - |A \cap B|$ |
| Intersection ($A \cap B$) | Elements common to both set A and set B. In this problem, this is the students playing both Chess and Badminton (11). | $|A \cap B|$ |
| Only in A | Elements in set A but not in set B. In this problem, this is students playing only Chess (17). | $|A| - |A \cap B|$ or $|A \cup B| - |B|$ |
| Only in B | Elements in set B but not in set A. In this problem, this is students playing only Badminton (17). | $|B| - |A \cap B|$ or $|A \cup B| - |A|$ |
| Total (when everyone is in A or B or both) | Sum of those only in A, only in B, and in both. | $|A \cup B| = (\text{Only A}) + (\text{Only B}) + (\text{Both})$ |
Additional Information: Solving Participation Problems
Problems like this, involving groups participating in different activities, are common examples of applications of basic set theory principles. While the problem can be solved with simple arithmetic as shown above, visualising it with a Venn diagram can also be very helpful, especially for more complex problems with three or more sets.
- Venn Diagrams: Circles representing each set (Chess, Badminton) overlap. The overlapping region is the intersection (both games). The parts of the circles outside the overlap represent those playing only that specific game. The sum of all distinct regions gives the total number of participants when everyone is included in at least one set.
- Inclusion-Exclusion Principle: The formula $|A \cup B| = |A| + |B| - |A \cap B|$ is known as the Principle of Inclusion-Exclusion. It accounts for the fact that simply adding the total number in Chess ($|A|$) and the total number in Badminton ($|B|$) would count the students playing both ($|A \cap B|$) twice, so you subtract them once. In this problem, we used a simplified version because we were given "only" categories.
Understanding these fundamental concepts helps in solving a wide range of problems involving overlapping groups or categories.
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