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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## 42 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.

Actual meaning of n(aUb) means all elements in either a or b or both. This a clear statement in all ncert reference books,go nd verify.

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

n(M) clearly says elements of set M while n(MΠS’) means intersection of set M and set S’ more clearly it means elements common in set M and set S’.

Actually its true n(AUB)=All elements that are in set A and set B ,disregarding the number of times they appear.

For example

A=(1,2,3,4,5,7,8,9) and set B=(10,11,13,2,4,3,14)

AUB=(1,2,3,4,5,7,8,9,10,11,13,14)

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

here we have A be set of prime numbers between 4 and 16 so the set be X = {5,7,11,13}

SUBSETS with exactly two elements in above set X = {5,7},{5,11},{5,13},{7,11},{7,13},{11,13}

Exactly a right solution by Venu.

I think answer is 15

I think you are right. I want to explain my answer in simple way.

170+140+120=430

Total=320

430-320=110

Common=35+50+40=125

125-110=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

here the given set be {1,2,3,4,5,6,7,8,9,10,11,12,13}

the total number of elements are (n)=13

so,the possible subsets for the above universal set be =n!=13!

13*12*11*10*9*8*7*6*5*4*3*2*1=6227026800

Nos of subsets could be found using formula 2^n ,where n represents nos of elements in a given set whose subsets are to be found.

possible subsets of set with 13 elements=2^13=8,192

please any help? Out of 45 students questioned, 42 like mathematics or english or both, 27 students like maths and 22 like english.

1) who didn’t like any of the subjects

2)who like maths only

3)both maths and english

HERE

we have the total number of students questioned (U)=45

the number of students like either mathematics or english n(MuE)=42

the no. of students like mathematics n(M)=27

the no.of students like english n(E)=22

1)the no.of students didn’t like any subject= U-n(MuE) =45-42=3

3)the no.of students like both maths and english = x

n(MuE)=n(M)+n(E)-x

42=27+22-x

42=49-x

x=49-42=7

2)who like math only =n(M)-x =27-7=20

A survey Carried in one of the Kenyan counties reveals the following data concerning unemployment.out of the population of people working or seeking for employment.9%unemployed,11%black,30%younger than 18,1%unemployed black youth,5%unemployed younger than 18,7%employed black youth,23%non black youth employment.

Help me find the %of not black

what is this in set theory ‘ £’

Can you please help me on mathematical induction?

In a servey concerning the reading habits of college of education of ikere , mathematics owo students,the proportion of students who read in the library,and hostel were found to be : library on 20% , library but not in class 25%, library 10%, library 30%, hostel 50%, hostel and classroom 10% none of the three reading places 25% using set theory find if 300 students were interviewed (1) how many students read in the classroom (2) how many students read in the hostel,if and only if they did not read in the class?

45_42=3

3dont like any

(AUB)=A+B_(AnB)

42 =27+22_Ano

(AnA)=49_42

=7 like both

27_7=20like math only

Somebody to helb n(u)=230, n(A)=75, n(B) = 90, n(c) =90, n(AnB) = 25, n(AnC)=15 , n(AnBnC) = 10, n(AuBuC)©

SOLVE n(AnBnC²)

Can anyone help me find this.

A researcher in Kampala city interviewed 150 students in science, 70 were physics students, 50 were registered in chemistry, 90 were biology students, 30 registered in physics and chemistry, 20 registered in chemistry and biology, 30 registered in physics and biology, and 10 registered in chemistry, biology and physics.

Using Venn diagram, find the following:-

(I) The number of students registered in two subjects.

(II) The number of students registered in only one subject.

(III) The number of students registered in none of the three subjects.

I want a set that deals with ” more than”

Please help me out : In a school of engineering a polytechnic required all students to take physics, mathematics or English in a graduating class of 252students 90 students take physics, 72 students take mathematics 162 take english 1/3 of those that take physics take English, 1/4 of those that take mathematics take English and 1/2 of those that take mathematics take physics .

(I,) how many students take all three courses

(ii) how many students take two or more courses

(iii) what percentage of the students take exactly one course

(Iv) what is the value of N(M n P) and N(P n E)

Please I need help solving this question

in an examination, 60 candidates passed science or maths.if 6 passed both subjects and 9 more passed maths than science.find the number of candidates who passed in each subject

please help me solve this

out of a group of 20 children, 10 drnk tea, 9 drink coffe, and 7 drink mil, 4 drink tea and coffe but none drinks both tea and milo.

find

a) how many drink milo and coffe

b) how many drink only coffe?

Hi can anyone help me

In a group of students , 38 enjoy video games , 12 enjoy movies and 24 enjoy solving math problems.8 students like all three activities , while 30 only like one of them.How many students like only two of the three activities.

Help me to solve this, In a class of 94 students taking maths, Bio and Chem equal number of students are taking only two subjects. The number of students taking maths, Bio, and Chem are 40, 35 and 38 respectively.Seven students are taking maths and Bio. Find

a. The number of students taking all courses

b. the number of students taking maths Bio and Chem

c. The number of students taking Bio and Chem

For me it’s a question please:

In an shs 1 ,of st Paul academy, 22 students take one or more of chemistry, economic, and government.12 takes econs (E),8 takes gov’t (G), and 7 takes chemistry (C). nobody takes econs and chem.And 4 students take econs and gov’t.

1. Using not and letter and veled above.write down the two statements mathematically in the left sentence.

[10/31, 11:26 PM] Fasila: 2. How many students take both chemistry and government?.

3. Government only.

in a class of 174 students, it was funds that. 37 students like biscuits, 60 students like chewing gum, and 111 students like sweets, 29 students like biscuits and sweets, 50 like chewing gum and sweets, and 13 students like chewing gum and biscuits. if 45 students neither like biscuits nor chewing gum nor sweets. how many students like chewing gum and biscuits but not sweets.

Each student in a class of 40 place at least one of the games hockey, football,& cricket. 16 play H, 20 play F, 26 play C ,5 H&F,14 F&C, 2H,C&F. Find the number of students who play H&C but not F.

Can anyone help me