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.

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.

__Solution__:

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

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

No. of students who played at least 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 – 15 – 20 – 15 + 10

Total number of students = 100.

Let a denote the number of people who played cricket and volleyball only.

Let b denote the number of people who played cricket and hockey only.

Let c denote the number of people who played hockey and volleyball only.

Let d denote the number of people who played all three games.

Accordingly, d = n (CnHnV) = 10

Now, n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Therefore, a = 15 – 10 = 5 [cricket and volleyball only]

b = 15 – 10 = 5 [cricket and hockey only]

c = 20 – 10 = 10 [hockey and volleyball only]

No. of students who played only cricket = n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

No. of students who played only hockey = n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

No. of students who played only volley ball = n(V) – [a + c + d] = 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.

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

Answer 1

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

Answer 2

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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Got queries unrelated to this article? Post them on our General Queries Page.## 15 Comments

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

Total number of students = 100

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

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.

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

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

n(MnS’)=n(M-S)

What about red and green only not blue

Even i believe that the answer is 100.

According to the theory we have to ADD overlapping regions, as well, to the individual regions in order to get TOTAL. So we have found Individual players as : 30 + 25 + 15 = 70. Now we have to add overlapping regions of the Venn diagram but carefully, we should not add them twice. So now overlapping regions are: 5(Crik and Hockey) + 5(Crick and VB) +10(VB and Hockey) = 20.

Adding these 3 overlapping has covered almost all the part of Venn diagram except the ALL. They still haven’t been taken into account. So 20 + ALL = 20 + 10 = 30.

Add to Individual Players : 70 + 30 = 100

i surely got the same answers as given.thanks alot for your guidelines

Can anyone help me and solve this question. In a bank of 320 staffs, 120 speak French, 140 speak English, 170 speak Arabic, 50 speak both French and English, 35 speak both English and Arabic, 40 speak both French and Arabic. Required ; (a) Determine the number of staffs who speak all the three languages. Thanks

n(U) = 320, //n(F)= 120, n(E)= 140, n(A)= 170,// n(F n E)= 50, n(E n A)= 35, n(F n A)= 40// n(F n E n A)=?.// So let n(F n E n A)= X.

THE FORMULA STATED THAT

n(U) = n(F) + n(E) + n(A) – n(F n E) – n(F n A) – n(E n A) + n(F n E n A) where n(F n E n A) is X

320 = 120 + 140 + 170 – 50 -35 + 40 + X

320 = 430 – 125 + X

320 =305 + X

320 – 305 =X

15 = X

X = 15, Where X is the number of staffs who spokes all the three languages.

So the number of staffs who spokes all the three subject are 15.

Any body please help me to solve this problem

If A = {X:X is a prime number between 4 and 16} list the subset with exactly two elements

I think answer is 15

only ten of staff speak both three languages since

/FUEUA/=/F/+/ E /+/E/-(/FnE/+/FnA/+/AnE/)+/FnEnA/

320=(120+140+170)-(50+40+30)+/FnEnA/

In a set {1,2,3,4,5,6,7,8,9,10,11,12,13} is a universal set, what are all the possible subset of the universal set