Cardinal Numbers - Meaning, Examples, Sets

Consider a teacher taking attendance in a class. The number of students present in the class is the cardinal number of the set of all students in the class. The cardinal number refers to the size or the count of the set. These concepts are broadly used in situations where two sets are compared. This concept of cardinal numbers of a set has broader applications in data analysis.

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This Story also Contains

1. Cardinal Number of Set
2. Cardinal Numbers of Sets Formula
3. Difference between cardinal and ordinary numbers
4. Solved Examples Based On the Cardinal Number of Sets

In this article, we will cover the concept of the cardinal number of sets. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of eleven questions have been asked on this concept, including one in 2019, four in 2020, one in 2021, and five in 2023.

Cardinal Number of Set

A set is simply a collection of distinct objects, considered as a whole. These objects, called elements or members of the set, can be anything: numbers, people, letters, etc. Sets are particularly useful in defining and working with groups of objects that share common properties. It is a well-defined collection of distinct objects and it is usually denoted by capital letters A, B, C, S, U, V, etc. Now let's look what is cardinal number in maths.

What is Cardinal Number of Set?

Cardinal numbers help us to count the number of things or people in or around a place or a group.

Take the example of a cricket team. There are eleven players in a cricket team, if we have to figure out the answer to the question. Currently, the number 11 (eleven) is a cardinal number.

The number of distinct elements in a finite set $A$ is called the Cardinal number of set $A$.

Cardinal number symbol: The cardinal number symbol of a set $A$ is $n(A)$

For example, if set $A=\{1,3,7,11,13\}$ then $n(A)=5$

Given, any two finite sets $A$ and $B$, then the Number of Elements in the union of sets $A \& B$ is given by

$
n(A \cup B)=n(A)+n(B)-n(A \cap B)
$

$
\text { If }(A \cap B)=\varphi \text {, then } n(A \cup B)=n(A)+n(B)
$

Given $A, B$, and $C$ are any finite sets, then the Number of Elements in the union of sets $A, B \& C$ is given by

$
n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(A \cap C)+n(A \cap B \cap C)
$

Now let us look into the cardinal number of some sets examples.

Cardinal Number of Some Sets Examples

The cardinal number of some sets,

1. Let $A = \{1,2,6,7,4,3,5,2,8\}$, then $n(A) = 9$

2. Let $B = \{a,d,c,t,r,v\}$, then $n(B) = 6$

Cardinal Numbers of Sets Formula

• For 2 disjoint sets, $n(AB)=n(A)+n(B)$

• For 2 overlapping sets, $n(AB)=n(A)+n(B)–n(AB)$

• For 3 disjoint sets, $n(ABC)=n(A)+n(B)+n(C)$

• For 3 overlapping sets, $ n(ABC)=n(A)+n(B)+n(C)–n(AB)–n(BC)–n(CA)+n(ABC)$

Important Notes:

• The collection of all the ordinal numbers can be denoted by the cardinal.
• Cardinal numbers can be written as words such as one, two, three, etc.
• Cardinal numbers tell how many items, whereas ordinal numbers show position or ranking.

Difference between cardinal and ordinary numbers

Cardinal numbers and ordinal numbers are both types of numbers used in mathematics and counting. Cardinal numbers represent the quantity or amount of objects or elements in a set, while ordinal numbers indicate the position or order of objects in a sequence. For example, the cardinal number "three" represents the quantity of three objects, while the ordinal number "third" indicates the position of an object in a sequence. While cardinal numbers are used for counting and measuring, ordinal numbers are used for ranking and ordering.

Recommended Video Based on the Cardinal Number of Sets

Solved Examples Based On the Cardinal Number of Sets

Example 1: If $U$ is the universal set, $n(A)=50, n(B)=60$, and $n(A \cap B)=30$. The total elements in the universal set are 200. Find $n\left(A^{\prime} \cap B^{\prime}\right)$

Solution:

$
\begin{aligned}
& \mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \\
& =50+60-30=80 \\
& n\left(A^{\prime} \cap B^{\prime}\right)=n(A \cup B)^{\prime}=200-80=120
\end{aligned}
$

Hence, the answer is 120.

Example 2: Given $n(A)=50, n(B)=30$ and $n(A U B)=x$ such that $n(A \cap B)=$
20. Find $x$

Solution:

We know,

$
\begin{aligned}
& n(A \cup B)=n(A)+n(B)-n(A \cap B) \\
& x=50+30-20=60
\end{aligned}
$

Hence, the answer is 60 .

Example 3: The sum of integers from 1 to 100 that are divisible by 2 or 5 is

Solution:

As we have learned
Number of Elements in Union A \& B -

$
\begin{aligned}
& n(A \cup B)=n(A)+n(B)-n(A \cap B) \\
& \text { - wherein }
\end{aligned}
$

Given $A$ and $B$ are any finite sets. then the Number of Elements in union $A \& B$ is given by this formula.

$
\begin{aligned}
& S=2+4+5+6 \\
& =(\text { sum of integers divisible by } 2)+(\text { sum of integers divisible by } 5)-(\text { sum of integers divisible by } 10(5 \times 2)) \\
& =(2+4+6+8 \ldots \ldots . .100)+(5+10+15 \ldots \ldots .100)-(10+20+\ldots \ldots \ldots+10) \\
& \text { sum of } n \text { term of an } A P \\
& =\frac{n}{2}(a+l) \\
& a=\text { first term } \\
& l=\text { last term } \\
& =\frac{50}{2}(2+100)+\frac{20}{2}(5+100)-\frac{10}{2}(10+100) \\
& =2550+1050-550=3050
\end{aligned}
$
Hence, the answer is 3050.

Example 4:

$\begin{equation}
\text { If } A \cap B=\phi, n(A)=50, n(B)=70 \text {. Then evaluate } n(A \cup B) \text {. }
\end{equation}$

Solution:

$\begin{equation}
\begin{aligned}
&\text { We know, }\\
&\begin{aligned}
& n(A \cup B)=n(A)+n(B)-n(A \cap B) \\
& \text { Since } A \cap B=\phi, n(A \cap B)=0 \\
& n(A \cup B)=50+70-0=120
\end{aligned}
\end{aligned}
\end{equation}$

Hence, the answer is 120.

.Example 5: In a class of 140 students numbered 1 to 140, all even-numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course, and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is;

Solution:

Number of Elements in Union A, B \& C -

$
\begin{aligned}
& n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(A \cap C)-n(B \cap C)+n(A \\
& \cap B \cap C) \\
& \text { - wherein }
\end{aligned}
$

Given $\mathrm{A}, \mathrm{B}$, and C are any finite sets. then the Number of Elements in union A , $B$ \& $C$ is given by this formula.

From the concept,
Let $n(M)=$ no. of students opted maths $=70$
$n(P)=$ student opted physics $=40$
$\mathrm{n}(\mathrm{C})=$ student opted chemistry $=28$
$n(M \cap P)=$ number of students opted for physics and maths $=23$
$n(M \cap C)=$ number of students opted for maths and chemistry = 14
$n(P \cap C)=$ number of students opted for physics and chemistry $=9$
$n(M \cap P \cap C)=$ number of students opted for all three students $=4$
So the total number of students who opted for at least one subject $=$

$
n(M \cup P \cup C)=n(M)+n(P)+n(C)-n(M \cap P)-n(M \cap C)-n(P \cap C)+n(M \cap P
$

$\cap$ C)
So putting the values, we have

$
n(M \cup P \cup C)=70+46+28-23-14-9+4=102
$

Hence total no. of students who have not adopted any course $=$ total number of students -total number of students who opted for at least one course $=140-$ $102=38$

Hence, the answer is 38.

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