Cardinal Numbers - Meaning, Examples, Sets

In real-life instances, the number referred to as the cardinal of a set is useful in finding the measure or the number of items in a given set or in a particular population. Basic to many tasks in everyday practice like stock control, resource apportionment, or data analysis, the cardinal number is immensely valuable. That is why it is critical to learn the cardinal number of a set because it helps to make the right decisions, reduce the number of errors in some processes, and even manage each element needed in our lives.

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.

Set

Sets are a foundational concept in mathematics, central to various fields such as statistics, geometry, and algebra. 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…...

What are cardinal numbers?

The numbers that respond to the query "How many" are known as cardinal numbers.
Take the example of a cricket team. There are eleven (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.
Cardinal numbers are also those that are used to indicate a specific location for an object, such as the position of students taking an exam in class or the positions of persons standing in a queue.

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.

Cardinal number of sets

The number of distinct elements in a finite set $A$ is called the Cardinal number of set $A$ and it is denoted by $n(A)$.

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)
$

Summary

The notion of the cardinal number of a set is going to help determine the number of elements and control their presence in different sets in practical everyday tasks. From the range of students in a classroom, books on a shelf as well as the cars in a parking lot, identification of the cardinal number enables one to decide on the most appropriate course of action not forgetting the aspect of division of resources. This understanding is universal education, commerce, logistics, and data processing, among various other disciplines.

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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: $\square$

$
\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.