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Sets - Cardinality of a Set-advanced

Grade 9CBSE

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

🔑Concepts

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Cardinality of a set AA, denoted by n(A)n(A), represents the number of distinct elements present in the set.

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A set is called a finite set if its cardinality is a whole number nn. If a set has no end to its elements, it is called an infinite set and its cardinality is not a finite number.

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The Principle of Inclusion-Exclusion for two sets AA and BB states that the number of elements in the union is the sum of elements in each set minus the elements in their intersection.

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For three sets A,B,A, B, and CC, the cardinality of the union involves adding individual cardinalities, subtracting double intersections, and adding back the triple intersection.

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If A∩B=∅A \cap B = \emptyset, the sets are disjoint, and n(A∪B)=n(A)+n(B)n(A \cup B) = n(A) + n(B).

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The power set P(A)P(A) contains all possible subsets of AA. If n(A)=nn(A) = n, then n(P(A))=2nn(P(A)) = 2^n.

📐Formulae

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

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−B)=n(A)−n(A∩B)n(A - B) = n(A) - n(A \cap B)

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

n(P(A))=2n(A)n(P(A)) = 2^{n(A)}

n(A′)=n(U)−n(A)n(A') = n(U) - n(A)

💡Examples

Problem 1:

In a group of 100100 people, 7070 can speak English and 4545 can speak Hindi. If every person speaks at least one of the two languages, find how many people can speak both English and Hindi.

Solution:

Let EE be the set of people who speak English and HH be the set of people who speak Hindi. Given: n(E∪H)=100n(E \cup H) = 100 n(E)=70n(E) = 70 n(H)=45n(H) = 45 We use the formula: n(E∪H)=n(E)+n(H)−n(E∩H)n(E \cup H) = n(E) + n(H) - n(E \cap H) 100=70+45−n(E∩H)100 = 70 + 45 - n(E \cap H) 100=115−n(E∩H)100 = 115 - n(E \cap H) n(E∩H)=115−100n(E \cap H) = 115 - 100 n(E∩H)=15n(E \cap H) = 15

Explanation:

We apply the inclusion-exclusion principle for two sets. Since everyone speaks at least one language, the union of the two sets equals the total number of people. Subtracting the union from the sum of individual sets gives the overlap (those who speak both).

Problem 2:

If set A={x:x is a prime factor of 210}A = \{x : x \text{ is a prime factor of } 210\}, find the number of elements in the power set of AA.

Solution:

First, find the prime factors of 210210: 210=2×3×5×7210 = 2 \times 3 \times 5 \times 7 So, set A={2,3,5,7}A = \{2, 3, 5, 7\}. The cardinality of set AA is: n(A)=4n(A) = 4 The number of elements in the power set P(A)P(A) is given by: n(P(A))=2n(A)n(P(A)) = 2^{n(A)} n(P(A))=24=16n(P(A)) = 2^4 = 16

Explanation:

The cardinality of the power set is always 2n2^n, where nn is the number of elements in the original set. Here, AA has 44 distinct prime factors.

Problem 3:

In a survey of 6060 students, 2525 play Cricket, 3030 play Football, and 2424 play Hockey. 1010 play Cricket and Football, 99 play Cricket and Hockey, 1212 play Football and Hockey, and 55 play all three games. Find how many students play none of the three games.

Solution:

Let C,F,C, F, and HH represent Cricket, Football, and Hockey respectively. n(C)=25,n(F)=30,n(H)=24n(C) = 25, n(F) = 30, n(H) = 24 n(C∩F)=10,n(C∩H)=9,n(F∩H)=12n(C \cap F) = 10, n(C \cap H) = 9, n(F \cap H) = 12 n(C∩F∩H)=5n(C \cap F \cap H) = 5 Using the formula for three sets: n(C∪F∪H)=25+30+24−(10+9+12)+5n(C \cup F \cup H) = 25 + 30 + 24 - (10 + 9 + 12) + 5 n(C∪F∪H)=79−31+5n(C \cup F \cup H) = 79 - 31 + 5 n(C∪F∪H)=53n(C \cup F \cup H) = 53 Total students n(U)=60n(U) = 60. Students playing none: 60−537\begin{array}{r} 60 \\ - 53 \\ \hline 7 \end{array} So, 77 students play none of the games.

Explanation:

We first calculate the number of students who play at least one game using the inclusion-exclusion principle for three sets, then subtract this value from the total number of students surveyed to find those who play none.