Counting Techniques Final (16-1-21)

Counting Techniques

Manoj V. Deshmukh M.Sc. Mathematics, SET,

Assistant Professor, PDEA’s Annasaheb Waghire College,

Otur, Pune.

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Dedicated to

My Father Late Bapu and Aai

PREFACE The counting techniques are useful in the development of several important concepts in

mathematics and computer science. An attempt has been made to cover basic and advance concepts in

each topic to care the needs of students who may earlier not have knowledge of these topics. My

purpose was to present this book in a precise readable manner with the concept. Each chapter contains

sufficiently large number of solved problems and illustrations to explain definition, principles and

theorems.

The book is designed for students of undergraduate and postgraduate levels for development of

more advanced mathematical concepts required in counting. It is also used for competitive

examinations for references.

While writing this book, I have benefited immensely by referring many books and publications.

I express my gratitude to all authors, publishers and many of them are listed in the references. If any

source has been left out by mistake, I seek their pardon.

I wish to express my sincere thanks to Shri Ajitdada Pawar – President, Shri Rajendraji Ghadge –

Vice President, Adv. Shri Sandeepji Kadam – Secretary and all honourable members of Pune District

Education Association for their best wishes and support.

I would like to express my sincere infinity thanks to my teachers Dr. N.K. Thakare, Dr. S.A.

Katre, Dr. T.T. Raghunathan, Dr. B.N. Waphare, Dr. Amir Athavale, Dr. Vilas Kharat, Prof.

Balasaheb Chavan, Prof. J.G. Chavan, Prof. Sonawane and many more who are enrolled in my life.

I wish to express my sincere thanks to Principals of various colleges in our institute in order as

Dr. M.G. Chaskar, Dr. N.L. Ghorpade, Dr. B.N. Zaware, Dr. P.N. Shelke, Dr. T.N. Salve,

Dr. Sharmila Chaudhary and Dr. Sushma Bhosale of Pune District Education Association for their best

wishes and support.

I would like to thank my dear all friends for their uninterrupted help and suggestions. I would

like to thank the editorial and publisher team of Himalaya Publishing House Pvt. Ltd.

Unlimited blessing from my mother Sharada, brothers Suresh Dada, Bandu Dada, Nitin Dada,

Pravin Dada and Nilesh, wife Manasi and children Prapti and Harshvardhan also deserve sincere

thanks for the unflagging support and encouragement they gave me while I worked on this book.

Although every care has been taken to check mistakes and misprints, yet it is difficult to claim

perfection. I am sure that errors remain to be found. I will gratefully accept any corrections, criticism

and valuable suggestions from the readers for the improvement of the book.

Manoj V. Deshmukh

CONTENTS

CHAPTER 1 SETS 3 – 26

1.1 Introduction

1.2 Sets

1.3 Subsets

1.4 Venn Diagrams

1.5 Basic Operations on the Sets

CHAPTER 2 FUNCTIONS 28 – 64

2.1 Introduction

2.2 Function

2.3 Graph of a Function

2.4 Composition of Functions

2.5 Injective, Surjective and Bijective Function

2.6 Increasing and Decreasing Function

2.7 Inverse Function

2.8 Floor and Ceiling Function

CHAPTER 3 SEQUENCES 67 – 74

3.1 Introduction

3.2 Sequence

3.3 Difference between Set and Sequence

3.4 Types of Sequence, Arithmetic and Geometric Sequence

CHAPTER 4 GENERATING FUNCTION 76 – 89

4.1 Introduction

4.2 Generating Function

4.3 Ordinary Generating Function

4.4 Basic Operation on Generating Functions

4.5 Exponential Generating Function

CHAPTER 5 RECURRENCE RELATION 91 – 124

5.1 Introduction

5.2 Modeling with Recurrence Relation

5.3 Homogeneous and Non-homogeneous Recurrence Relation

5.4 Methods of Solving Recurrence Relation

5.5 Solution of Recurrence Relation by Generating Function

CHAPTER 6 BASIC COUNTING 126 – 203

6.1 Factorial Notation

6.2 Introduction

6.3 Basic Counting Principles

6.4 Permutations

6.5 Combinations

6.6 Generating functions for Counting

6.7 Permutations with Repeated Objects

6.8 Combinations with Repeated Objects

6.9 Partitions and Binomial Coefficient

CHAPTER 7 ADVANCED COUNTING TECHNIQUES 205 – 221

7.1 Introduction

7.2 Inclusion-Exclusion Principle

7.3 Applications Inclusion-Exclusion Principle

References

List of Symbols

Srinivasa Ramanuja (December 22, 1887, Erode, India — April 26, 1920, Kumbakonam).

He was an Indian mathematician whose contributions to the theory of numbers include

pioneering discoveries of the properties of the partition function. Though he had almost no formal

training in pure mathematics, he made substantial contributions to mathematical analysis, number

theory, infinite series, and continued fractions, including solutions to mathematical problems then

considered unsolvable. Ramanujan initially developed his own mathematical research in isolation.

When he was 15 years old, he obtained a copy of George Shoobridge Carr’s Synopsis of

Elementary Results in Pure and Applied Mathematics, 2 vol. (1880–86). This collection of thousands

of theorems, many presented with only the briefest of proofs and with no material newer than 1860,

aroused his genius. Having verified the results in Carr’s book, Ramanujan went beyond it, developing

his own theorems and ideas. In 1903, he secured a scholarship to the University of Madras but lost it

the following year because he neglected all other studies in pursuit of mathematics.

Ramanujan continued his work, without employment and living in the poorest circumstances.

After marrying in 1909 he began a search for permanent employment that culminated in an interview

with a government official, Ramachandra Rao. Impressed by Ramanujan’s mathematical prowess, Rao

supported his research for a time, but Ramanujan, unwilling to exist on charity, obtained a clerical post

with the Madras Port Trust.

In 1911, Ramanujan published the first of his papers in the Journal of the Indian Mathematical

Society. His genius slowly gained recognition, and in 1913 he began a correspondence with the British

mathematician Godfrey H. Hardy that led to a special scholarship from the University of Madras and a

grant from Trinity College, Cambridge. Overcoming his religious objections, Ramanujan travelled to

England in 1914, where Hardy tutored him and collaborated with him in some research.

In England, Ramanujan made further advances, especially in the partition of numbers (the

number of ways that a positive integer can be expressed as the sum of positive integers; e.g., 4 can be

expressed as 4, 3 + 1, 2 + 2, 2 + 1 + 1, and 1 + 1 + 1 + 1). His papers were published in English and

European journals, and in 1918 he was elected to the Royal Society of London. In 1917, Ramanujan

2 Counting Techniques 2

had contracted tuberculosis, but his condition improved sufficiently for him to return to India in 1919.

He died the following year, generally unknown to the world at large but recognized by mathematicians

as a phenomenal genius, without peer since Leonhard Euler (1707–83) and Carl Jacobi (1804–51).

Ramanujan left behind three notebooks and a sheaf of pages (also called the “lost notebook”)

containing many unpublished results that mathematicians continued to verify long after his death.

Sets 3

1.1 INTRODUCTION Set is the fundamental concept in the theory of Discrete Structures. The theory of sets was first

introduced by the German Mathematician G. Cantor (1845-1918) who defined set theory as a

collection of objects. The language of the sets has become an important tool for all branches of

Mathematics.

1.2 SET – DEFINITION A set is a well-defined collection of the objects. The individual objects in the collection is called

member or element of the set.

Set Notation

The set denote by upper case letters of English alphabet, viz., A, B, C…and to denote the objects

in it. Let A be any set. If an element x belong to the set A then we denoted it by x A, read as x is an

element of the set A or x is member of the set A. If an element x not belong to the set A then we

denoted it by x A, read as x is not a element of the set A or x is not a member of the set A. The

symbol stands for belongs to.

If we consider the set P of positive prime integers then 2 P but 4 P.

Examples

1. The collection of first 10 positive integers.

2. The collection of past prime ministers of India.

3. The collection of topper students of the college.

The sets are usually denoted by capital alphabets as A, B, C, X, Y, Z etc.

1 SETS


Following are the basic sets (Standard notations) denote by common letters as:

N: The set of all natural numbers.

R: The set of all real numbers.

IR+: The set of all positive real numbers.

Z: The set of all integers

Z+: The set of all positive integers.

Q: The set of all rational numbers.

Q : The set of all non-zero rational numbers.

Q+: The set of all positive rational numbers.

C: The set of all complex numbers.

C : The set of all non-zero complex numbers.

+, Q

+, IR

+ represents the positive elements of the sets Z, Q, IR respectively.

A set can be specified in two different ways as:

1. Roster method (Tabular Form)/Listing Method: A set is represented by listing all the

elements by commas within braces {}.

For example:

1. The set of all natural numbers N = {1,2,3,…}

2. The set of all prime numbers = {2,3,5,7,11,…}

3. The set of all even integers = { … –4, –2,0,2,4,…}

4. Set of all vowels in the English alphabets = {a,e,i,o,u}

2. Rule Method (Set-builder form): A set whose elements represented by stating the property or

properties they must have to be members.

For example:

Consider the following set

A = {1,2,3,…10} which is in roster/tabular form. It is represented in set builder form as –

A = {x/x is a natural number less than or equal to 10} or {x N/x 10}

X = {1,3,5,…} (Roster method)

X = {x/x is an old number} (Rule method)

3. Statement form: The set describing by statement only.

For example:

1. Set of positive integers.

2. Set of all even integers.

Sets 5

Universal Set

If all sets considered during a specific discussion are subsets of a given set, then this set is called

as Universal set and is denoted by U. It is also referred to as the universe of discourse.

For example, In a discussion involving the sets X = {1,2,3,4}, Y = {2,5,6,9} and Z = {1,3,7} so

here one may choose U = {1,2,3,4,…9} as a universal set.

Empty Set

A set has no element then it is known as empty set or null set. It is denoted by the symbol . i.e.,

= { }.

For example:

1. A set of odd numbers between 3 and 5.

2. Set of students got above 100 marks out of 50 marks in Mathematics.

Problem 1: Give an example of a set which is empty.

Solution: The easiest way consider nice thing as a set X = {n N/n2 + 1 = 0}, Since n is natural

number and clearly there is no natural number n which satisfy the equation n2 + 1 = 0; hence, X is

empty set.

Problem 2: What about the set X = {n IR/n2 + 1 = 0}?

Solution: Here, n is real number, the equation n2 + 1 = 0 satisfies only when n

2 = –1 which is

impossible. Hence, again given set X is empty.

Singleton Set

A set consist of only one element, such set known as singleton set.

For example:

1. A = {2}

2. Set of integers between 7 and 9.

3. Set of even number between 11 and 13.

Another example is a set X = {n N/n2 – 1 = 0}, for n = 1 equation n

2 – 1 = 0 satisfies, therefore

X = {1} it is singleton set.

Finite set: A set X is said to be finite if it has exactly n distinct elements where, n is non-negative

integers. i.e., X contains only finite number of elements.

For example:

1. A = {10,11,12,13,14,15,100}

2. B = {x N/x 10}

3. Set of people in the world


Infinite set: A set is not finite then it is known as infinite set.

For example:

1. Set of stars in the Galaxy.

2. Set of points on a straight line.

3. The set of positive integer greater than 10.

Equality of two sets: Two sets A and B are said to be equal, if both having same elements Or

Every element of A is also element of B and vice versa. i.e., A B and B A. We denote it by

simply A = B.

For example:

1. The sets A = {1,2,3} and B = {3,1,2} are equal because they have the same elements.

2. The set A = {1,3,3,3,4} and B = {1,3,4} also equal sets since, they having same elements.

Note: The order in which the elements listed in the set is not important.

1.3 SUBSET The set A is said to be a subset of the set B if and only if every elements of the set A is also an

elements of set B. It is denoted as A B read as A is subset or equal to B.

i.e., if x A then x B.

For Example: Set of even integers is subset of set of integers.

Clearly, N Z+ Z Q IR C, where C 13 set of complex numbers.

Let A = {0,1,2,3}, B = {0,1,2,3,4,5} and C = {2,4} then clearly A B, C B but C A.

Following statement are easy to verify:

Note:

1. Every set is a subset of itself.

2. The empty set is a subset of any set.

1.4 VENN DIAGRAM Sometimes diagrams or pictures are very helpful to understand or thinking. Swiss mathematician

Leonard Euler first of all gave an idea to represent a set by the points in a closed curve. Later on

British mathematician John Venn (1834-1923) brought this idea in practice, therefore such diagrams

are called as Venn diagrams. A Venn diagram is a pictorial representation of sets in which the

universal set U is represented by a rectangle and the sets within U by circles. (See Fig. 1.1) i.e., Sets

can be represented by the points enclosed within rectangle and its subsets are represented by the points

enclosed by the circles within the rectangles. To understand the set theory concepts, the Venn

diagrams are very important for beginners.

Sets 7

Let U be universal set, A and B are subsets such that B A U, C U and D U.

Fig. 1.1

Clearly, From above Venn diagram B is subset of A, C is not subset of A because some part of C

outside from set A i.e., some elements of C not in A, D is not subset of A.

Problem 3: Write all subsets of the set A = {a, b, c}.

Solution: The required subsets are , {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}.

Problem 4: If A = {a,b,c, {a,b}, {{a,b}}} then which of the following is true or false?

(i) b A (ii) a A

(iii) {a,b} A (iv) {{a,b}} A

Solution:

(i) b A is true,because b is an element of the set A.

(ii) a A is False, because a is an element of A not subset.

(iii) {a,b} A is true, because {a,b} is an element of set A.

(iv) {{a,b}} A, is true, because {{a,b}} is subset of A.

Proper Subsets

A set X is said to be a proper subset of set Y, if

(i) X is subset of Y i.e., X Y

(ii) X is not equal to Y i.e., X Y.

i.e., A set X is said to be a proper subset of set Y, if X is subsets of Y and if there exist at least

one element of Y which is not in X.

For example: If X = {1,2,3,4,5} and Y = {1,2,3,4,5,6,7,8,9,10} then X is proper subset of Y.

U

A

B

D

C


Problem 5: Let A = {x, y, z}. Which of the following are subsets of A? Which are proper subsets

of A?

(a) P = {y}

(b) Q = {y, z, x}

(c) R = {z, y}

(d) S = {k, x, y}

Solution:

(a) Clearly, the element y of P is also element of set A therefore P A, here P A thus P is

proper subset of A.

(b) The all elements of Q are also elements of set A, thus Q A or Q = A.

(c) All element of set R are also element of set A but R A, therefore R is proper subset of A.

i.e., R A.

(d) The element k in S but it is not the member of set A, therefore here S A.

Theorem: Every set is a subset of itself.

Proof: Let A be any set. Every element of set A is always member of A itself, therefore A is

subset of A itself, i.e., A A.

Note: Empty (Null) set is always subset of every set.

Power Set

Let A be a given set. The set of all subsets of set A is known as power set of A and it is denoted

by the symbol (A).

For example:

1. Let A = {a, b} be a given set. Its possible subsets are , {a}, {b}, {a, b}.

Thus, power set (A) = { , {a}, {b}, {a, b}}.

2. Let A = {a, b, c} be a given set.

Its possible subsets are , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}.

Thus, power set (A) = { , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} }.

Note: ( ) = {}.

Theorem: If a set A has n elements then power set has 2 subsets of A i.e., | (A)| = 2 .

Proof: Suppose a set A has n elements, we have to prove it has 2 subsets. Use mathematical

induction on n N.

Step-I (Basis of Induction): let n = 1 i.e., the set A has one element, so it has only two subsets

and A itself. Hence, there are 2 subsets are possible.

Sets 9

Step-II (Induction step): Assume that set A has n 1elements then it has 2 subsets are

possible.

Now, To prove if the set A has n elements then it has 2n subsets.

Suppose a set A has n elements as A = { , , … , , }, partitioned the set A as

= { { , , … , }, }.

Therefore, |A| = n. Let S = {a1, a2, …, an – 1}

Then |S| = n 1, so S has 2 subsets, according to the inductive hypothesis.

A = { }.

Clearly, every subset of S is also a subset of A. Any subset of A either contains the element an or

it doesn’t contain an.

Case I: If a subset of A doesn’t contain, an then it is also a subset of S, and S has total 2 of

those subsets.

Case II: On the other hand, if a subset of A does contain the element a, then that subset is formed

by including an in one of the 2 subsets of S,

Thus A has 2 subsets containing an.

From this discussion, we conclude that A has 2 subsets containing an, and another 2

subsets not containing an. Hence, total number of subsets of A is 2 + 2 = (2) 2 = 2

Thus, A set has n elements then it has 2 subsets are possible.

Problem 6: List all the elements of the following sets:

(a) { / , < 12}

(b) {x/x N and x is prime and x < 25}

(c) {x/x Z and x is a root of x + 5x + 4 = 0}

Solution:

(a) The given set contains all integers whose square less than 12. Therefore, possible value of x

are = 0, ±1, ±2, ±3

{ / , < 12} = {0, ±1, ±2, ±3}

(b) The given set contains all natural numbers less than 25 and are prime numbers. Therefore,

possible value of x are = 2,3,5,7,11,13,17,19,23

{x/x N and x is prime and x < 25} = {2,3,5,7,11,13,17,19,23}

(c) The given set contains all integers such that they are solution of given equation

x + 5x + 4 = 0

(x + 1)(x + 4) = 0

x = 1, x = 4


Therefore, possible values of x are = 1, 4

{x/x Z and x is a root of x + 5x + 4 = 0} = { 1, 4}

1.5 BASIC OPERATIONSW ON THE SETS

Union of Two Sets

Let A and B are two non-empty sets .The union of A and B is the set of common and uncommon

elements of both sets and it is denoted by A B ,the symbol is used for union. For example, If

A = {1,2,3,4} and B = {2,4,5,9} then A B = {1,2,3,4,5,9}.

Union of two sets shown by shaded region in the given Venn diagram.

Fig. 1.2

Problem 7: If A B then, what about the A B?

Solution: A B = B

Intersection of Two Sets

Let A and B are two sets. The intersection of two sets A and B is the set of only common

elements of both sets and it is denoted by A B, the symbol is used for intersection.

For example, If A = {1,2,3,4} and B = {2,4,5,9} then A B = {2,4}.

Intersection of two sets shown by shaded region in the given Venn diagram.

U

B

A

A U B

Sets 11

Fig. 1.3

Problem 8: If A B then, what about the A B?

Solution: A B = A.

Problem 9: If A = {n N/n2 – 4 = 0} and B = {n N/n is even prime}

then what is A B and A B?

Solution: A = {n N/n2 – 4 = 0} = {2}

B = {n N/n is even prime} = {2}

Thus, A B = {2} and A B = {2}.

Generalization of Union and Intersection

Let us consider finite collection of n sets, say A , A , A , … , A (n > 1). In this case, we write the

union as:

A = A A … A = {x/x A for some i, 1 i n}

A = A A … A = {x/x A for each i, 1 i n}

Disjoint Sets

If two non-empty sets A and B have no element in common in both then they are said to be

disjoint sets. Or two sets are said to disjoint if their intersection is empty. i.e., A B = .

For example:

1. If A = {1,3} and B = {2,4,5,9} then A B = .

2. If A = {n N/n2 – 1 = 0} and B = {n N/2n – 4 = 0} then A B = .

U

A B


Disjoint sets A and B shown in the given Venn diagram.

Fig. 1.4

Properties of Union and Intersection of Sets

1. If A and B are any two sets, then A B = B A. (Commutative Rule)

Proof: We have to prove A B = B A it is enough to prove A B B A and B A A B.

Part 1: To prove A B B A

Let x be any arbitrary element in A B.

x A B

x A or x B

x B or x A

x B A

Hence, A B B A … (1)

Part 2: To prove B A A B

Let x be any arbitrary element in B A.

x B A

x B or x A

x A or x B

x A B

Hence, B A A B … (2)

From (1) and (2)

A B = B A

A

U

B

Sets 13

2. If A and B are any two sets, then A B = B A. (Commutative Rule)

Proof: We have to prove A B = B A it is enough to prove A B B A and B A A B.

Part 1: To prove A B B A

Let x be any arbitrary element in A B.

x A B

x A and x B

x B and x A

x B A

Hence, A B B A … (1)

Part 2: To prove B A A B

Let x be any arbitrary element in B A.

x B A

x B and x A

x A and x B

x A B

Therefore,

B A A B … (2)

From (1) and (2) A B = B A

3. If A, B and C are any three sets, then (A B) C = A (B C). (Associative Rule)

Proof: We have to prove (A B) C = A (B C)

it is enough to prove (A B) C A (B C) and A (B C) (A B) C.

Part 1: To prove (A B) C A (B C)

Let x be any arbitrary element in (A B) C.

x A B or x C

x A or x B or x C

x A or( x B or x C)

x A (B C)

Hence, (A B) C A (B C) … (1)

Part 2: To prove A (B C) (A B) C

Let x be any arbitrary element in A (B C).

x A or x (B C)


x A or x B or x C

x A B or x C x (A B) C

Hence, A (B C) (A B) C … (2)

From (1) and (2) (A B) C = A (B C)

4. If A, B and C are any three sets, then (A B) C = A (B C). (Associative Rule)

Proof: We have to prove (A B) C = A (B C)

it is enough to prove (A B) C A (B C) and A (B C) (A B) C.

Part 1: To prove (A B) C A (B C)

Let x be any arbitrary element in (A B) C.

x (A B) C

x A B and x C

x A and x B and x C

x A and( x B and x C)

x A (B C)

Hence, (A B) C A (B C) … (1)

Part 2: To prove A (B C) (A B) C

Let x be any arbitrary element in A (B C).

x A (B C)

x A and x (B C)

x A and x B and x C

(x A and x B)and x C

x (A B) and x C

x (A B) C

Hence, A (B C) (A B) C … (2)

From (1) and (2) (A B) C = A (B C).

5. If A is any set, then A A = A (Idempotent Rule)

Proof: We have to prove A A = A, it is enough to prove A A A and A A A.

Let x be any arbitrary element in A A

x A A

x A or x A

x A

Hence, A A A and A A A implies that A A = A.

Sets 15

6. If A is any set, then A A = A (Idempotent Rule)

Proof: We have to prove A A = A, it is enough to prove A A A and A A A.

Let x be any arbitrary element in A A

x A A

x A and x A

x A

Hence, A A A and A A A implies that A A = A.

7. If A and B are any two sets, then (i) A (A B) = A ii) A (A B) = A. (Absorption Rule)

Proof: Let x be any arbitrary element in A (A B)

x A (A B)

x A or x (A B)

x A and x B

x A

Hence, A (A B) A … (1)

Now consider x A x A (A B) A A (A B) … (2)

From (1) and (2)

A = A (A B)

Similar proof for A (A B) = A.

8. If A and B are any two sets, then A A B and B A B

Proof: Let us prove A A B. Consider x be an arbitrary element in A

x A

x A or B

Hence, A A B

Similarly, B A B.

9. If A and B are any two sets, then A B A B, A B A, and A B B.

Proof: Let us prove A B A B.

Consider x be an arbitrary element in A B

x A B

x A and x B

x A or x B

x A B


Hence, A B A B

Similarly, A B A, and A B B.

10. If A is any set then A = A.

11. If A is any set then A = .

12. If A is any subset of universal set U then A U =U.

13. If A is any subset of universal set U then A U =A.

14. If A is subset of B i.e., A B then A B = A and if B A then B A = B.

Cardinality of a Set: Let A be a finite set with n elements. The cardinality of A is the number of

elements in the set A and it is denoted as |A| and |A| = n.

For example:

1. X = { a, e, i, o, u} then |X| = 5.

2. X = Set of all English alphabets, then |X| = 26.

3. A = {1,2,3, … ,100} then |A| = 100.

Note: 1. If = then |A| = 0.

2. If then |A| |B|.

Problem 10: If A = {n Z/n 1 = 0}, what is the value of |A|?

Solution: |A| = 2

Theorem: If A and B are any two disjoint finite sets then |A B| = |A| + |B|.

Proof: Suppose A and B are any two disjoint sets as A = { , , … , } and B = { , , … , }

|A| = m, |B| = n

Since A and B are disjoint i.e., A B = |A B| = 0.

Here, A = { , , … , , , , … , }

|A B| = m + n

|A B| = |A| + |B|

Corollary: Let A , A , … , A are any finite collection of mutually disjoint sets

i.e., A A = , for i then

A = | A |

Sets 17

Difference of Two Sets

Let A and B are two non-empty sets. The difference of two sets A and B is the set containing

elements of A, but not belongs to in B and it is denoted by A B or \ .

i.e.,

= { / , } = { / , }

For example: If A = {1,2,3,4} and B = {2,4,5,9} then A B = {1, 3} and B A = {5, 9}.

Thus, A B B A.

Difference of two sets shown by shaded region in the given Venn diagram.

Fig. 1.5 Fig. 1.6

Symmetric Difference of Two Sets

Let A and B are two sets. The symmetric difference of these two sets A and B is denoted by

A B or A B, is the set containing those elements in either A or B but not in both A and B.

A B = (A B)U(B A) or

A B = (A B) (A B)

A B = {x A B/x A B}

Symmetric difference of two sets shown by shaded region in the given Venn diagram.

U

A

U

B B

A

A-B B-A


( ) ( )

Fig. 1.7

Problem 11: If A = {1,2,3,4,5} and B = {3,5} then A B =?

Solution: A B = {x/x A but x B} = {1,2,4}

Problem 12: If A is the set of all positive prime integers between 1 to 100 and B is the set of all

positive odd integers between 1 to 100, then what is the set B A?, A – B?

Problem 13: If A B, then what is the set A B, B A?

Properties of Difference of Two Sets

1. Let A be any set then A A = .

Proof: Let x be any arbitrary element of A A

i.e., x A A x A and x A i.e., element x is in A and x not in A implies that such x

does not exists.

thus A A = .

2. Let A and B are any two sets, then A B A also B A B.

Proof: Let x be any arbitrary element of A B

i.e., x A – B x A, x B x A

Thus, A B A. Similarly B A B.

3. A = A.

4. (A B) A = A.

Complement of Set

Let A be a subset of universal set U. The complement of A denoted by A or A and it is the set of

elements of U but not in A. i.e., A = {x U/x A} i.e., A = U A

Complement of set A shown by shaded region in the given Venn diagram.

B

U

A

Sets 19

Fig. 1.8

Problem 14: Let U be the universal set and A is subset of U as U = {x/1 < x < 20} and A be the

set of all positive prime integers between 1 to 20, What is the complement of A?

Solution: Here, U = {x/1 < < 20} and A be the set of all prime integers between 1 to 20, thus

A = {2,3,5,7,11,13,17,19}.

A = {x U/x A}

A = {4,6,8,9,10,12,14,15,16,18}

Problem 15: Let U be he Universal set of all positive integers and A be the set of integers greater

than 5. Find the complement of A.

Solution: Given that the set A is the set of integers greater than 5. i.e.,

A = {x U/x > 5}

Thus, the complement of A is

A = {x U/x 5}

Properties of Complement of Set

(a) A A = U

(b) A A =

(c) (A ) = A

(d) U = and = U

(e) (D’ Morgan’s Rule) Let U be universal set and A, B are any two subsets of U then

1. (A B) = A B

2. (A B) = A B .

U

A


Proof:

(a) To prove A A = U

Let x A A i.e.,

x A or x A

x U

Thus, A A = U

(b) A A = (Do as exercise)

(c) To prove (A ) = A Let x (A ) x A x A , therefore (A ) A … (1)

Similarly let x A x A x (A ) therefore A (A ) … (2)

from (1) and (2)

(A ) A and A (A )

Hence, (A ) = A.

(d) U = and = U (Obvious proof, check!)

(e) (1) (A B) = A B

To prove (A B) = A B it is enough to prove (A B) A B and A B

(A B)

Let x be an arbitrary element in (A B)

i.e., x (A B)

x A B

x A and x B

x A and x B

x A B

Hence, (A B) A B and A B (A B)

Therefore, (A B) = A B .

(2) (A B) = A B .

To prove (A B) = A B it is enough to prove (A B) A B and A B

(A B)

Let x be an arbitrary element in (A B)

i.e., x (A B)

x A B

x A or x B

x A or x B

Sets 21

x A B

Hence, (A B) A B and A B (A B)

Therefore, (A B) = A B .

Distributive Laws

Let A, B and C are any three sets then

(a) A (B C) = (A B) (A C)

(b) A (B C) = (A B) (A C)

Proof:

(a) To prove A (B C) = (A B) (A C)

Let x be an arbitrary element in A (B C) i.e.,

x A (B C)

x A or x (B C)

x A or (x B and x C)

(x A or x B) and (x A or x C)

x (A B)and x (A C)

x (A B) (A C)

Hence, A (B C) = (A B) (A C).

(b) To prove A (B C) = (A B) (A C)

Let x be an arbitrary element in i.e.,

x A (B C)

x A and x (B C)

x A and (x B or x C)

(x A and x B) or (x A and x C)

x (A B)or x (A C)

x (A B) (A C)

Hence, A (B C) = (A B) (A C)

Problem 16: Let A, B and C denote the subsets of universal set U and C denote the complement

of C in U. A C = B C and A C = B C , then prove that A = B.

Solution: Let U be the universal set

A = A U

= A (C C )


= (A C) (A C ) [Distributive Law]

= (B C) (B C ) [by given condition]

= B (C C ) [Distributive Law]

= B U

Therefore, A = B

Problem 17: Let A, B and C are subsets of a set X, if (A C) (B C ) = then prove that

A B = , where C is the complement of C in the set X.

Solution: Given that (A C) (B C ) =

(A C) = and (B C ) = … (1)

Since (B C ) = B C =

There are three cases i) B = ii) B = C iii) B C

Case (i) If B = then A B = .

Case (ii) If B = C then A B = A C = . From (1)

Case (iii) If B C then A B A C = . From (1) Hence Proved.

Problem 18: Let A = {1,2,3,4,5,6} and B = {0,1,4,5} then find

(i) A B (ii) A B (iii) A B (iv) B A.

Solution:

(i) A B = {x/x A x B} = {0,1,2,3,4,5,6}

(ii) A B = {x/x A and x B} = {1,4,5}

(iii) A B = {x/x A x B} = {0,2,3,6}

(iv) B A = {x/x B x A} = {0}

Ordered Pairs

Let X and Y be two non-empty sets and x X and y Y then the ordered pairs of elements x and

y, denoted by (x, y).

Clearly, (x, y) (y, x) whenever x y.

Theorem: The ordered pairs (x, y) = (z, w) if and only if x = z and y = w, for x, z X and

y, w Y.

Cartesian Product of Two Sets

Let X and Y be two non-empty sets and x X and y Y then the cartesian product of X and Y

denoted by X × Y and is the set of all possible ordered pairs of elements x and y.

X × Y = {(x, y)/x X and y Y}

For example, If X = {1,2,3} and Y = {a, b} then Cartesian product of X and Y is

Sets 23

X × Y = {(x, y)/x X and y Y}

X × Y = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}

Y × X = {(y, x)/y Y and x X}

Y × X = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}

As we observe these two sets X × Y and Y × X is not equal i.e., X × Y Y × X.

Note: If |X| = m and |Y| = n then | X × Y| = mn.

Problem 19: Consider three sets, P = {x N/3 < x < 9}, Q = {x N/–1 < x < 5} and

R = {x N/0 < x < 9} are subsets of universal set U, U = {x /–2

(a) (i) P (ii) Q (iii) R

(b) (i) P R (ii) P R (iii) Q R (iv) R P

(c) Verify that (i) (P R ) = P R (ii) P (P Q) = P

Solution: Given that universal set U = { 2, 1,0,1,2,3,4,5,6,7,8,9,10}

P = {4,5,6,7,8} , Q = {0,1,2,3,4} and R = {1,2,3,4,5,6,7,8} are subsets of U

(a) (i) P = {x U/x P} = { 2, 1,0,1,2,3,9,10}

(ii) Q = {x U/x Q} = { 2, 1,5,6,7,8,9,10}

(iii) R = {x U/x R} = { 2, 1,0,9,10}

(b) (i) P R = {1,2,3,4,5,6,7,8}

(ii) P R = {4,5,6,7,8}

(iii) Q R = {0,1,2,3,4,5,6,7,8}

(iv) R P = {1,2,3}

(c) (i) Now to verify (P R ) = P R

We know P R = {1,2,3,4,5,6,7,8}

(P R ) = { 2, 1,0,9,10} … (1)

P = {x U/x P} = { 2, 1,0,1,2,3,9,10}

R = {x U/x R} = { 2, 1,0,9,10}

P R = { 2, 1,0,9,10} … (2)

From (1) and (2)

(P R ) = P R

(ii) Now to verify P (P Q) = P

Since (P Q) = {4}

P (P Q) = {4,5,6,7,8} = P.


Problem 20: Let U = {1,2,3,4,5,6,7,8,9}, A = {1,2,3,7}, B = {4,5,6,7}, C = {1,3,6}, D = {6,8,9}

Compute the following:

(i) , (ii) , (iii) , (iv) , (v) , (vi) , (vii) ,

(viii) , (ix) ( ), (x) ( ), (xi) , (xii) , where A is complement of A.

Solution: Given that

U = {1,2,3,4,5,6,7,8,9}, A = {1,2,3,7}, B = {4,5,6,7}, C = {1,3,6}, D = {6,8,9}

A = {4,5,6,8,9}, B = {1,2,3,8,9}, C = {2,4,5,7,8,9}, D = {1,2,3,4,5,7}

(i) A B = {7}

(ii) B C = {4,5,7}

(iii) A C = {2,4,5,6,7,8,9}

(iv) A D = {4,5}

(v) A C = {4,5,8,9}

(vi) B D = {1,2,3,4,5,6,7}

(vii) A B C = {1,2,3,4,5,6,7}

(viii) A B C = {}

(ix) A (B C) = {1,3,7}

(x) A (B C) = {1,2,3,6,7}

(xi) A B = (A B) (A B) = {1,2,3,4,5,6}

(xiii) A C = (A C) (A C) = {2,6,7}

Problem 21: Determine the sets A and B, given that

A B = {1,2,4,5,7,8,9}, A B = {1,2,4}, B A = {7,8}

Solution:

= ( ) ( ) = { , , , , }

Fig. 1.9

U

B

A

Sets 25

= ( ) ( ) = { , , , }

Fig. 1.10

Problem 22: Determine the sets A and B, given that

A B = {1,2,4,5,7,8,9,10}, A B = {2,4,7} and A B = {1,8}

Solution:

A = (A B) (A B) = {1,2,4,7,8}

B = (A B) (A B) = {2,4,5,7,9,10}

EXERCISE 1. List the members of the sets.

(a) {x/x is a real number such that x = 1}

(b) {x/x is a positive integer less than 15}

(c) {x/x is odd prime < 10}

2. Find the cardinality of each of the following sets.

(a) A = {a}

(b) B = {{a}}

(c) C = {a, {a}, a, {a} }

(d) D =

3. Find the power set of each of the following sets.

(a) {a} (b) {1,2,3,4,}

U

B

A


4. Let A = {a, b, c, d, e, f} and B = {b, d, e, f, g, h} then find

(i) A B (ii) A B

(iii) A B (iv) B A

5. Let A = { 2, 1,0,1,2, … ,10}, B = {0,1,2, … ,8} and C= {3,4,6,10} then find

(i) (A B) C (ii) (A B) C

(iii) (A B) C

6. Find the set A and B, if A B = {1,5,7,8}, B A = {2,10} and A B = {3,6,9}.

7. Draw the Venn diagram for the sets A and B such that

(i) A B = A (ii) A B = A

(iii) A B = A (iv) A B = B A

(v) A B = B A

8. Find the symmetric difference of A = {1,3,5} and B = {1,2,3}.

9. What can you say about the sets A and B if A B = A?

10. Let X = {2,4}. which of the following sets are equal to A?

(a) A = {4,2}

(b) A = {x/x 6x + 8 = 0}

(c) A = {x/x 4 = 0 and x + 2 = 0}

Answers

1. (a) {–1,1}, (b) {1,2,…14}, (c){3,5,7}

2. (a) |A| = 1, (b) |B| = 1, (c) 3, (d) |D| = 0

3. (a) (A) = { , {a}}

(b) (A) = { , {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3},

{2,4}, {3,4}, {1,2,3}, {1,2,4}, {2,3,4}, {1,3,4}, A}

4. (i) A B = {a, b, c, d, e, f, g, h}, (ii) A B = {b, d, e, f}, (iii) A B = {a, c} , (iv) B A =

{g, h}

5. (i) A B C = { 2, 1, … ,10}, (ii) (A B) C = {3,4,6,10}, (iii) A B C = {3,4,6,10}

6. A = {1,3,5,6,7,8,9} and B = {2,3,6,9,10}

8. A B = {2,5}

10. (a) Yes, (b) Yes, (c) No.

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