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Sets - Venn Diagram-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Venn Diagram is a visual representation of the relationship between different sets. The Universal Set UU is usually represented by a rectangle, and its subsets are represented by circles or ovals inside it.

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The intersection of three sets A∩B∩CA \cap B \cap C is the region common to all three circles in a Venn diagram.

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The region 'Exactly one of the sets' refers to elements that belong to only one set and not the intersections. For set AA, this is A−(B∪C)A - (B \cup C).

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The region 'Exactly two of the sets' refers to elements in the intersections of two sets but excluding the triple intersection: (A∩B−C)∪(B∩C−A)∪(C∩A−B)(A \cap B - C) \cup (B \cap C - A) \cup (C \cap A - B).

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The Complement of a set A′A', denoted as U−AU - A, represents all elements in the universal set that are not in AA. In a Venn diagram, this is the area outside circle AA but inside the rectangle UU.

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The Principle of Inclusion-Exclusion for three sets is used to find the total number of elements in the union n(A∪B∪C)n(A \cup B \cup C) by accounting for overlaps.

📐Formulae

n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B) samples

n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)

n(A′)=n(U)−n(A)n(A') = n(U) - n(A)

n(Exactly two sets)=n(A∩B)+n(B∩C)+n(C∩A)−3n(A∩B∩C)n(\text{Exactly two sets}) = n(A \cap B) + n(B \cap C) + n(C \cap A) - 3n(A \cap B \cap C)

n(Exactly one set)=n(A)+n(B)+n(C)−2[n(A∩B)+n(B∩C)+n(C∩A)]+3n(A∩B∩C)n(\text{Exactly one set}) = n(A) + n(B) + n(C) - 2[n(A \cap B) + n(B \cap C) + n(C \cap A)] + 3n(A \cap B \cap C)

💡Examples

Problem 1:

In a group of 100100 students, 4545 study Mathematics, 4040 study Physics, and 3535 study Chemistry. 1515 study Math and Physics, 1212 study Physics and Chemistry, 1010 study Math and Chemistry, and 55 study all three subjects. How many students study none of the three subjects?

Solution:

Let MM be the set of students studying Mathematics, PP for Physics, and CC for Chemistry. We are given: n(U)=100n(U) = 100 n(M)=45n(M) = 45, n(P)=40n(P) = 40, n(C)=35n(C) = 35 n(M∩P)=15n(M \cap P) = 15, n(P∩C)=12n(P \cap C) = 12, n(M∩C)=10n(M \cap C) = 10 n(M∩P∩C)=5n(M \cap P \cap C) = 5 Using the inclusion-exclusion formula: n(M∪P∪C)=45+40+35−(15+12+10)+5n(M \cup P \cup C) = 45 + 40 + 35 - (15 + 12 + 10) + 5 n(M∪P∪C)=120−37+5n(M \cup P \cup C) = 120 - 37 + 5 n(M∪P∪C)=88n(M \cup P \cup C) = 88 Number of students studying none of the subjects: n(U)−n(M∪P∪C)=100−88=12n(U) - n(M \cup P \cup C) = 100 - 88 = 12

Explanation:

First, we find the total number of students who study at least one subject using the formula for the union of three sets. Then, we subtract this from the total number of students in the universal set to find those who study none.

Problem 2:

Using the data from the previous example (n(M)=45,n(P)=40,n(C)=35,n(M∩P)=15,n(P∩C)=12,n(M∩C)=10,n(M∩P∩C)=5n(M)=45, n(P)=40, n(C)=35, n(M \cap P)=15, n(P \cap C)=12, n(M \cap C)=10, n(M \cap P \cap C)=5), find how many students study exactly two subjects.

Solution:

The number of students studying exactly two subjects is given by the formula: n(Exactly two)=[n(M∩P)−n(M∩P∩C)]+[n(P∩C)−n(M∩P∩C)]+[n(C∩M)−n(M∩P∩C)]n(\text{Exactly two}) = [n(M \cap P) - n(M \cap P \cap C)] + [n(P \cap C) - n(M \cap P \cap C)] + [n(C \cap M) - n(M \cap P \cap C)] Substituting the values: n(Exactly two)=(15−5)+(12−5)+(10−5)n(\text{Exactly two}) = (15 - 5) + (12 - 5) + (10 - 5) n(Exactly two)=10+7+5n(\text{Exactly two}) = 10 + 7 + 5 n(Exactly two)=22n(\text{Exactly two}) = 22

Explanation:

To find students studying exactly two subjects, we take the intersections of each pair of sets and subtract the triple intersection from each, as the triple intersection represents students studying three subjects, not exactly two.