**Cardinal Number of a set**

The number of distinct elements or members in a finite set is known as the cardinal number of a set. Basically, through cardinality, we define the size of a set. The cardinal number of a set A is denoted as n(A), where A is any set and n(A) is the number of members in set A.

Consider a set A consisting of the prime numbers less than 10.

Set A ={2, 3, 5, 7}.

As the set A consists of 4 elements, therefore, the cardinal number of set A is given as n(A) = 4.

## Properties related to difference, union and intersection and the cardinal number of set

**i) Union of Disjoint Sets:**

If A and B are two finite sets and if A ∩ B = ∅, then

n(A ∪ B) = n(A) + n(B)

In simple words if A and B are finite sets and these sets are disjoint then the cardinal number of Union of sets A and B is equal to the sum of the cardinal number of set A and set B.

Figure 1- Disjoint sets

The union of the disjoint sets A and B represented by the Venn diagram is given by A ∪ B and it can be seen that A ∩ B = ∅ because no element is common to both the sets.

**ii) Union of two sets:**

If A and B are two finite sets, then

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Simply, the number of elements in the union of set A and B is equal to the sum of cardinal numbers of the sets A and B, minus that of their intersection.

Figure 2- Union of two sets

In the figure given above the differently shaded regions depict the different disjoint sets i.e. A – B, B – A and A ∩ B are three disjoint sets as shown and the sum of these represents A ∪ B. Hence,

n (A ∪ B) = n (A – B) + n(B – A) + n(A ∩ B)

**iii) Union of three sets**

If A, B and C are three finite sets, then;

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

This is clearly visible from the Venn diagram that the union of the three sets will be the sum of the cardinal number of set A, set B, set C and the common elements of the three sets excluding the common elements of sets taken in pairs of two.

Figure 3-Union of three sets

## Video Lesson

### Applied Concept – Cardinality of Sets

### Solved Example

Let us see an example to make our point clear.

**Example: There is a total of 200 students in class XI. 120 of them study mathematics, 50 students study commerce and 30students study both mathematics and commerce. Find the number of students who**

**i) Study mathematics but not commerce**

**ii) Study commerce but not mathematics**

**iii) Study mathematics or commerce**

**Solution:** The total number of students represents the cardinal number of the universal set. Let A denote the set of students studying mathematics and set B represent the students studying commerce.

Therefore,

n (U) = 200

n(A) = 120

n(B) = 50

n(A ∩ B) = 30

The Venn diagram represents the number of students studying mathematics and commerce.

i) Here, we are required to find the difference of sets A and B.

n(A) = n(A – B) + n(A ∩ B)

n(A-B) = n(A) – n(A ∩ B)

⇒ n (A – B) = 120 – 30 = 90

The number of students who study mathematics but not commerce is 90.

ii) Similarly here, we are required to find the difference of sets B and A

n (B) = n (B – A) + n (A ∩ B)

⇒ n (B – A) = 50 – 30 = 20

The number of students who study commerce but not mathematics is 20.

iii) The number of students who study mathematics or commerce

n (A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ n(A ∪ B) = 120 + 50 – 30 = 140

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