Set theory has its own notations and symbols that can seem unusual for many. In this tutorial, we look at some solved examples to understand how set theory works and the kind of problems it can be used to solve.

#### Definition

A set is a collection of objects.

It is usually represented in flower braces.

For example:
Set of natural numbers           = {1,2,3,…..}
Set of whole numbers             = {0,1,2,3,…..}

Each object is called an element of the set.

The set that contains all the elements of a given collection is called the universal set and is represented by the symbol ‘µ’, pronounced as ‘mu’.

For two sets A and B,

• n(AᴜB) is the number of elements present in either of the sets A or B.
• n(A∩B) is the number of elements present in both the sets A and B.
• n(AᴜB) = n(A) + (n(B) – n(A∩B)

For three sets A, B and C,

• n(AᴜBᴜC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C)

Consider the following example:

Question: In a class of 100 students, 35 like science and 45 like math. 10 like both. How many like either of them and how many like neither?

Solution:

Total number of students, n(µ) = 100

Number of science students, n(S) = 35

Number of math students, n(M) = 45

Number of students who like both, n(M∩S) = 10

Number of students who like either of them,

n(MᴜS) = n(M) + n(S) – n(M∩S)

→ 45+35-10 = 70

Number of students who like neither = n(µ) – n(MᴜS) = 100 – 70 = 30

The easiest way to solve problems on sets is by drawing Venn diagrams, as shown below.

As it is said, one picture is worth a thousand words. One Venn diagram can help solve the problem faster and save time. This is especially true when more than two categories are involved in the problem.

Let us see some more solved examples.

Problem 1: There are 30 students in a class. Among them, 8 students are learning both English and French. A total of 18 students are learning English. If every student is learning at least one language, how many students are learning French in total?

Solution:

The Venn diagram for this problem looks like this.

Every student is learning at least one language. Hence there is no one who fall in the category ‘neither’.

So in this case, n(EᴜF) = n(µ).

It is mentioned in the problem that a total of 18 are learning English. This DOES NOT mean that 18 are learning ONLY English. Only when the word ‘only’ is mentioned in the problem should we consider it so.

Now, 18 are learning English and 8 are learning both. This means that 18 – 8 = 10 are learning ONLY English.

n(µ) = 30, n(E) = 10

n(EᴜF) = n(E) + n(F) – n(E∩F)

30 = 18+ n(F) – 8

n(F) = 20

Therefore, total number of students learning French = 20.

Note: The question was only about the total number of students learning French and not about those learning ONLY French, which would have been a different answer, 12.

Finally, the Venn diagram looks like this.

Problem 2: Among a group of students, 50 played cricket, 50 played hockey and 40 played volley ball. 5 played both cricket and hockey, 10 played both hockey and volley ball, 5 played cricket and volley ball and 10 played all three. If every student played at least one game, find the number of students and how many played only cricket, only hockey and only volley ball?

Solution:

n(C) = 50, n(H) = 50, n(V) = 40

n(C∩H) = 5

n(H∩V) = 10

n(C∩V) = 5

n(C∩H∩V) = 10

No. of students who played atleast one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 5 – 10 – 5 + 10

Total number of students = 130.

No. of students who played only cricket = n(C) – [n(C∩H) + n(C∩V) + n(C∩H∩V)] = 50 – (5+5+10) = 30.

No. of students who played only hockey = n(H) – [n(C∩H) + n(H∩V) + n(C∩H∩V)] = 50 – (5+10+10) = 25.

No. of students who played only volley ball = n(V) – [n(H∩V) + n(C∩V) + n(C∩H∩V)]=40-(10+5+10) = 15.

Alternatively, we can solve it faster with the help of a Venn diagram.

The Venn diagram for the given information looks like this.

Subtracting the values in the intersections from the individual values gives us the number of students who played only one game.

## Set Theory Quiz: Solve these problems for practice

Problem 1

In a group, there were 115 people whose proofs of identity were being verified. Some had passport, some had voter id and some had both. If 65 had passport and 30 had both, how many had voter id only and not passport?

A. 30
B. 50
C. 80
D. None of the above

B.

Explanation

Let us draw the Venn diagram for the given information.

n(PᴜV) = n(P) + n(V) – n(P∩V)

115 = 65+n(V) – 30

n(V) = 80

People with only voter id = 80-30 = 50

Problem 2

Among a group of people, 40% liked red, 30% liked blue and 30% liked green. 7% liked both red and green, 5% liked both red and blue, 10% liked both green and blue. If 86% of them liked at least one colour, what percentage of people liked all three?

A. 10
B. 6
C. 8
D. None

C.

Explanation:

n(RᴜBᴜG) = n(R) + n(B) + n(G) – n(R∩B) – n(B∩G) – n(R∩G) + n(R∩G∩B)

86 = 40+30+30-5-10-7+ n(R∩G∩B)

Solving this gives 8.

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1. kris says:

In the problem 2, the way i understand sets is that those that played all three are included in those that played both cricket and hockey. Such that inorder to obtain the number of those that played cricket and hockey but not volleyball, you’ll have to subtract the number of those who played all 3 from the number of those that played cricket hockey. This is just my thought. I don’t know what you think

• Sandeep says:

Total number of students = 100

• Younus says:

I have solved it but the no. Is 130 you can verify again.

2. Younus says:

I have an objection with one of your formula statement which states:
n(AᴜB) is the number of elements present in either of the sets A or B.
But in the example given bellow you consider elements of both sets and add them up this completely violates the above statement which tells add only elements from either of the two sets.

3. Seun says:

In your example 1. How will you differentiate n(M) and n(M n S’). That is, maths and maths only.

4. Vidhani says:

n(M) =n(M-S) +n(MnS)

n(MnS’)=n(M-S)