Direction: Consider the following for the next 02 (two) items that follow: In a school, all the students play at least one of three indoor games – chess, carom and table tennis. 60 play chess, 50 play table tennis, 48 play carom, 12 play chess and carom, 15 play carom and table tennis, 20 play table tennis and chess.
What can be the maximum number of students in the school?
123
Finding the Maximum Number of Students Using Set Theory
The problem asks for the maximum number of students in a school where every student plays at least one of three indoor games: chess, carom, and table tennis. We are given the number of students playing each game individually and the number playing combinations of two games.
This problem can be solved using the Principle of Inclusion-Exclusion for three sets.
Understanding the Given Data
Let's denote the sets of students playing chess, carom, and table tennis as $C$, $Ca$, and $T$ respectively. The given information is:
- Number of students playing chess, $|C| = 60$
- Number of students playing table tennis, $|T| = 50$
- Number of students playing carom, $|Ca| = 48$
- Number of students playing chess and carom, $|C \cap Ca| = 12$
- Number of students playing carom and table tennis, $|Ca \cap T| = 15$
- Number of students playing table tennis and chess, $|T \cap C| = 20$
Since all students play at least one game, the total number of students in the school is the number of students in the union of the three sets, i.e., $|C \cup Ca \cup T|$.
Applying the Principle of Inclusion-Exclusion
The formula for the union of three sets is:
\(|C \cup Ca \cup T| = |C| + |Ca| + |T| - (|C \cap Ca| + |Ca \cap T| + |T \cap C|) + |C \cap Ca \cap T|\)
Calculating the Union with Known Values
Substitute the given values into the formula:
\(|C \cup Ca \cup T| = 60 + 48 + 50 - (12 + 15 + 20) + |C \cap Ca \cap T|\)
\(|C \cup Ca \cup T| = 158 - (47) + |C \cap Ca \cap T|\)
\(|C \cup Ca \cup T| = 111 + |C \cap Ca \cap T|\)
Here, $|C \cap Ca \cap T|$ represents the number of students who play all three games.
Maximizing the Number of Students
To find the maximum number of students in the school, we need to maximize the value of $|C \cap Ca \cap T|$.
The number of students who play all three games must be less than or equal to the number of students who play any pair of these games. That is:
- $|C \cap Ca \cap T| \le |C \cap Ca| = 12$
- $|C \cap Ca \cap T| \le |Ca \cap T| = 15$
- $|C \cap Ca \cap T| \le |T \cap C| = 20$
To maximize $|C \cap Ca \cap T|$, we must choose the largest possible value that satisfies all these conditions. The largest value that is less than or equal to 12, 15, and 20 is the minimum of these values, which is 12.
So, the maximum possible value for $|C \cap Ca \cap T|$ is 12.
Calculating the Maximum Total Number of Students
Now, substitute the maximum value of $|C \cap Ca \cap T| = 12$ back into the union formula:
Maximum \(|C \cup Ca \cup T| = 111 + 12\)
Maximum \(|C \cup Ca \cup T| = 123\)
Therefore, the maximum number of students in the school is 123.
Final Answer Summary
Using the principle of inclusion-exclusion and maximizing the intersection of all three sets, we found the maximum total number of students.
Given:
| Game/Combination | Number of Students |
|---|---|
| Chess (C) | 60 |
| Carom (Ca) | 48 |
| Table Tennis (T) | 50 |
| Chess and Carom (C ∩ Ca) | 12 |
| Carom and Table Tennis (Ca ∩ T) | 15 |
| Table Tennis and Chess (T ∩ C) | 20 |
Formula: \(|C \cup Ca \cup T| = |C| + |Ca| + |T| - (|C \cap Ca| + |Ca \cap T| + |T \cap C|) + |C \cap Ca \cap T|\)
Substitute values: \(|C \cup Ca \cup T| = 60 + 48 + 50 - (12 + 15 + 20) + |C \cap Ca \cap T|\)
\(|C \cup Ca \cup T| = 158 - 47 + |C \cap Ca \cap T|\)
\(|C \cup Ca \cup T| = 111 + |C \cap Ca \cap T|\)
Maximum value of $|C \cap Ca \cap T|$ is the minimum of the pairwise intersections: min(12, 15, 20) = 12.
Maximum \(|C \cup Ca \cup T| = 111 + 12 = 123\)
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Set Union (\(\cup\)) | Represents elements belonging to at least one of the sets. Total students here. |
| Set Intersection (\(\cap\)) | Represents elements belonging to all the sets considered. Eg., C ∩ Ca are students playing both Chess and Carom. C ∩ Ca ∩ T are students playing all three. |
| Principle of Inclusion-Exclusion | A counting technique to find the number of elements in the union of multiple sets by summing individual set sizes, subtracting pairwise intersections, adding back three-way intersections, and so on. For three sets A, B, D: \(|A \cup B \cup D| = |A| + |B| + |D| - (|A \cap B| + |B \cap D| + |A \cap D|) + |A \cap B \cap D|\). |
| Maximizing Union | In the Inclusion-Exclusion formula, the union is maximized when the intersection of all sets is maximized, given the other values are fixed. The maximum value of the three-set intersection is limited by the minimum of the pairwise intersections. |
Additional Information: Venn Diagrams for Set Problems
Problems like this can often be visualized using Venn diagrams. For three sets, you would draw three overlapping circles within a rectangle (representing the universal set, which is the school here). Each region in the Venn diagram represents a specific combination of games played (e.g., only chess, chess and carom but not table tennis, all three games, etc.).
Let x be the number of students playing all three games, \(|C \cap Ca \cap T| = x\).
- Students playing C and Ca ONLY: \(|C \cap Ca| - x = 12 - x\)
- Students playing Ca and T ONLY: \(|Ca \cap T| - x = 15 - x\)
- Students playing T and C ONLY: \(|T \cap C| - x = 20 - x\)
- Students playing C ONLY: \(|C| - (|C \cap Ca| - x) - (|T \cap C| - x) - x = 60 - (12 - x) - (20 - x) - x = 60 - 12 + x - 20 + x - x = 28 + x\)
- Students playing Ca ONLY: \(|Ca| - (|C \cap Ca| - x) - (|Ca \cap T| - x) - x = 48 - (12 - x) - (15 - x) - x = 48 - 12 + x - 15 + x - x = 21 + x\)
- Students playing T ONLY: \(|T| - (|Ca \cap T| - x) - (|T \cap C| - x) - x = 50 - (15 - x) - (20 - x) - x = 50 - 15 + x - 20 + x - x = 15 + x\)
The total number of students is the sum of students in all these regions, plus the region playing all three games ($x$).
Total = (28 + x) + (21 + x) + (15 + x) + (12 - x) + (15 - x) + (20 - x) + x
Total = (28 + 21 + 15 + 12 + 15 + 20) + (x + x + x - x - x - x + x)
Total = 111 + x
This confirms the formula result. To maximize the total, we maximize x. As established, x is maximum when it is the minimum of the pairwise intersections, which is 12.
Maximum Total = 111 + 12 = 123.
Important Questions from Operations on Sets
What is the number of natural numbers less than or equal to 1000 which are neither divisible by 10 nor 15 nor 25?
If A = {x ∈ R : x 2+ 6x - 7 < 0} and B = {x ∈ R : x 2+ 9x + 14 > 0}, then which of the following is/are correct?
1. (A ∩ B) = (-2, 1)
2. (A - B) = (-7, -2)
Select the correct answer using the code given below:A, B, C and D are four sets such that A ∩ B = C ∩ D = ϕ. Consider the following:
1. A ∪ C and B ∪ D are always disjoint.
2. A ∩ C and B ∩ D are always disjoint.
Which of the above statements is/are correct?A coin is tossed three times. Consider the following events:
A: No head appears
B: Exactly one head appears
C. At least two heads appear
Which one of the following is correct?
If C = { 2, 4, 6, 8, 10, 12, 14, 16 }, and D = {5, 10, 15, 20}, then the number of elements in the set D - C is:
