**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 cardinal number of set A is given as n(A) = 4.

**Properties related to difference of sets, union and intersection of sets and 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 cardinal number of set A and set B.

Figure 1- Disjoint sets

The union of the disjoint sets A and B as represented by the Venn diagram is given by A ∪ B and it can be seen that A ∩ B = ∅ as 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 put 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 different 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 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

Let us see an example to make our point clear.

**Example: There are 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 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.

\(~~~~~~~~~\)

n(A) = n(A – B) + n(A ∩ B)

⇒ n (A – B) = 120 – 30 = 90

\(~~~~~~~~~\)

\(~~~~~~~~~\)

n (B) = n (B – A) + n (A ∩ B)

⇒ n (B – A) = 50 – 30 = 20

\(~~~~~~~~~\)

\(~~~~~~~~~\)

\(~~~~~~~~~\)

⇒ n(A ∪ B) = 120 + 50 – 30 = 140

\(~~~~~~~~~\)

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