Download - The objectives of Discrete Mathematical Structures are · The objectives of Discrete Mathematical Structures are: • To introduce a number of Discrete Mathematical Structures (DMS)

The objectives of Discrete Mathematical Structures are:

• To introduce a number of Discrete Mathematical Structures (DMS) found to be serving

as tools even today in the development of theoretical computer science.

• Course focuses on of how Discrete Structures actually helped computer engineers to

solve problems occurred in the development of programming languages.

• Also, course highlights the importance of discrete structures towards simulation of a

problem in computer science and engineering.

• Introduction of a number of case studies involving problems of Computer

Technology.

Outcomes of this course are:

• A complete knowledge on various discrete structures available in literature.

• Realization of some satisfaction of having learnt that discrete structures are indeed

useful in computer science and engineering and thereby concluding that no mistake has

been done in studying this course.

• Gaining of some confidence on how to deal with problems which may arrive in

computer science and engineering in near future.

• Above all, students who studied this course are found to be better equipped in a

relative sense as far as preparation for entrance examinations involving placement

opportunities.

What is Discrete Mathematics then?

• Mathematics is broadly divided into two parts; (i) the continuous mathematics and

(ii) the discrete mathematics depending upon the presence or absence of the limiting

processes.

• In the case of continuum Mathematics, there do exists some relationship / linkage

between various topics whereas Discrete Mathematics is concerned with study of

distinct, or different, or un-related topics of mathematics curriculum; it embraces

several topical areas of mathematics some of which go back to early stages of

mathematical development while others are more recent additions to the discipline. The

present course restricts only to introducing discrete structures which are being used

as tools in theoretical computer science.

• A course on Discrete mathematics includes a number of topics such as study of sets,

functions and relations, matrix theory, algebra, Combinatorial principles and discrete

probability, graph theory, finite differences and recurrence relations, formal logic and

predicate calculus, proof techniques - mathematical induction, algorithmic thinking,

Matrices, Primes, factorization, greatest common divisor, residues and application to

cryptology, Boolean algebra; Permutations, combinations and partitions; Recurrence

relations and generating functions; Introduction to error-correcting codes; Formal

languages and grammars, finite state machines. linear programming etc. Also, few

computer science subjects such as finite automata languages, data structures, logic

design, algorithms and analysis were also viewed as a part of this course.

• Because of the diversity of the topics, it is perhaps preferable to treat Discrete

Mathematics, simply as Mathematics that is necessary for decision making in non-

continuous situations. For these reasons, we advise students of CSE / ISE / MCA, TE

(Telecommunication Engineering) to study this course, as they needs to know the

procedure of communicating with a computer may be either as a designer, programmer,

or, at least a user.

• Of course, in today’s situation, this is true for all, although we do not teach to students

of other branches of engineering. In some autonomous engineering colleges, DMS is

being offered s an elective. Considering these view points, you are informed to

undertake a course on Discrete Mathematical Structures so that you will be able to

function as informed citizens of an increasingly technological society.

• Also, Discrete Mathematics affords students, a new opportunity to experience success

and enjoyment in Mathematics classes. If you have encountered numerous difficulties

with computation and the complexities of Mathematics in the past, then may I say that

this course is soft and a study requires very few formal skills as prerequisites.

• In case if you are discouraged by the routine aspects of learning Mathematics, Discrete

Structures provides you a unique opportunity to learn Mathematics in a much different

way than the one employed by your teachers previously. Above all, Discrete

Mathematics is vital, exciting, and no doubt is useful otherwise you would not have

been suggested to register for this course.

• Further, Discrete Mathematics course serves as a gateway for a number of subjects in

computer science and engineering. With these motivations, here, we initiate a detailed

discussion on some of the topics: These include Basic set theory, Counting techniques,

Formal Logic and Predicate calculus, Relations and functions in CSE, Order relations,

Groups and Coding etc.

• Before, continuing, let me mention the difference between Discrete Mathematics and

other Mathematics; consider a bag of apples and a piece of wire. In the former, the

apples sit apart discretely from each other while in the latter, the points on a wire spread

themselves continuously from one end to the other.

• Thus, the numbers 0, 1, 2, 3, . . . are sufficient to handle DMS, where as a real variable

taking values continuously over a range of values is required to deal with continuum

Mathematics. Hence,

• Discrete Mathematics + Limiting Processes = Continuum Mathematics.

• Prescribed text book:

• Discrete and Combinatorial Mathematics by R. P. Grimaldi, PHI publications, 5th

edition (2004).

• Reference Books:

• Discrete Mathematical Structures by Kenneth Rosen, Tata McGraw Hill Publications

• Discrete Mathematical Structures by Kolman, Busby and Ross, PHI publications

Basic set theory

A set is a well defined collection of well defined distinct objects. A set is usually denoted

by using upper case letters like A, B, G, T, X etc. and arbitrary elements of the set are

denoted using lower case letters such as a, b, g, v etc.

Universal set: The set of all objects under some investigation is called as universe or

universal set, denoted by the symbol U.

Consider a set A. Let x be an element of A. This we denote symbolically by x AÎ .

On the other hand, if y is not an element of the set, the same is written as y AÏ . Thus,

it is clear that with respect to a set A, and an element of the universal set U, there are only

two types of relationship possible; (i) the element x under question is a member of the set

A or the element x need not be a member of A.

This situation may well be described by using binary numbers 0 and 1. We set :x 1 to

indicate that the element in question is a member of the set A. The notation :x 0

means that the element under study is not a member of the set A. There are a number of

ways of do this task. (i) Writing the elements of a set within the braces. For example,

consider A = {dog, apple, dead body, 5, Dr. Abdul Kalam, rose}. Certainly, A qualifies as a

set.

(ii) A set may be explained by means of a statement where elements satisfying some

conditions. Consider { | is a Engineering College Affiliated to VTU, Belgaum}V x x=

(iii) Definition of a statement may be given by means of a statement like Z denotes the set

of all integers. Thus, Z = {. . . -2, -1, 0, 1, 2, 3,. . }.

A null set is a one not having any elements at all. It is denoted by the symbol { } or as .

Give few examples of null set or empty set.

Compliment of a set: Let A be a set. The compliment of A is defined as a set containing

elements of the universe but not the elements of the set A. Thus, { | }A x U x A= Î Ï

Subset of a set: Let A and B be two sets. We say that A is a subset of B whenever B

contains all the elements of A or equivalently, each element of the set A is a member of

B. This is denoted by the symbol A BÍ . In a construction of a subset, we have the

option of including the null set as well as, the set itself.

Power set of a set: Let X be a set, then collection of all subsets of X is called as power

set of X, denoted by P(X). Thus, ( ) { | }P X A A X= Í . For example, if X = {a, b, c}, then

power set of X is given as ( ) { , { }, { }, { }, { , }, { , }, { , }, { , , }}P X a b c a b a c b c a b cf= . In

general if A set X has n elements, then its power set will have n2 elements. This is

because, during the construction of a subset of X using the n – elements of X, we have a

choice of either including an element of X in the subset or excluding the same element in

the subset. Thus, each of the n – elements of X has exactly 2 choices, therefore the total

number of choices will turn out to be n2 .

Proper subset of a Set: Let A and B be two sets. One says A is a proper subset of B if B

contains at least one element that is not in A. This is denoted by A BÌ .

Note: Difference between proper subset and a subset is the following:

• In a subset of X, we can find both the null set and the set X itself.

• In the case of a proper subset of X, we can include the null set but not the set X.

Example of a Discrete Mathematical Structure

Consider the universal set, U. Let { , , }X A B C= be a collection of subsets of U. Consider

the set operator “subset defined on { , , }X A B C= . We have . and A A AfÍ Í1

. If , , then , . If , then A B B C A C A B B AÍ Í Í Í Í2 3 . Thus, we can claim that

( ), X Í an example of a discrete structure.

Equal Sets: Let A and B be two sets. We say that A = B whenever both A and B have

exactly same elements. Equivalently if and A B B AÍ Í . .

Equivalent sets: Let A and

same number of elements. This is denoted by

called cardinal number of the set. This is denoted by

Union Operator: Consider a universal set Let

Then union of A and B is defined as

explained by using Venn diagram

The functioning of union operator may also be explained using membership table and Venn

diagram. The membership table is given below.

x : x A, BHere, we set up 1 to mean

From the membership table, it is clear that

all other instances, Î Èx A B

and B be two sets. We say A is equivalent to B

same number of elements. This is denoted by .A Bp The number of elements in a set is

of the set. This is denoted by |A|.

Discussion on Set operations

: Consider a universal set Let , , A B C be any three sets in

Then union of A and B is defined as { | or }A B x U x A x BÈ = Î Î Î The same can be

diagram and by employing a membership table.


diagram. The membership table is given below.

x : x A, BÎHere, we set up 1 to mean and x : x A, BHere, we set up 0 to mean that

A B ÈA B

1 1 1

1 0 1

0 1 1

0 0 0

From the membership table, it is clear that Ï Èx A B only when and Ï Ïx A x B

x A B .

U:U

A BÈ

if A and B have

number of elements in a set is

e any three sets in .U

The same can be


x : x A, BÏHere, we set up 0 to mean that .

and Ï Ïx A x B . For

A BÈ

Properties with respect to Union Operator:

1. (idempote

2. (Commutative la

3. ( ) ( ) (Associative law)

4. (Identity law)

5.

f f

È =

È = È

È È = È È

È = È =

È = È =

A A A

A B B A

A B C A B C

A A A

A U U A U (Universal law)

6. If , then Í È =A B A B B

Thus, we have another discrete structure, namely,

subsets of the universal set,

A set together with an operation and objects of

as discrete structure.

Intersection Operator: Consider a universal set Let

.U Then Intersection of A and B is defined

same can be explained by using Venn diagram and by employing a membership table.

The same well be explained by using membership table

Properties with respect to Union Operator:

1. (idempotent law)

2. (Commutative law)


4. (Identity law)

A B C A B C

(Universal law)

A B A B B

discrete structure, namely, ( ), , , UA B C where A, B,

subsets of the universal set, U. Before, continuing what is a discrete structure?

A set together with an operation and objects of the set satisfying some properties is called

: Consider a universal set Let , , A B C be any three sets in

of A and B is defined as { | and }Ç = Î Î ÎA B x U x A x B


The same well be explained by using membership table

A B ÇA B

1 1 1

1 0 0

0 1 0

0 0 0

U

ÇA B

A, B, and C are all

. Before, continuing what is a discrete structure?

the set satisfying some properties is called

e any three sets in

{ | and }Ç = Î Î ÎA B x U x A x B . The


Note: Here, only when and . For other instances, Î Ç Î Î Ï Çx A B x A x B x A B

clear that the two operators union and inter

view of this, these operators are

Properties with respect to Intersection Operator:

1. (idempote

2. (Commutative la


4. (Identity law)

5. f f f

Ç =

Ç = Ç

Ç Ç = Ç Ç

Ç = Ç =

Ç = Ç =

A A A

A B B A

A B C A B C

A U U A A

A A (Universal law)

6. If , then Í Ç =A B A B A

Thus, we have generated a discrete structure, namely,

all subsets of the universal set, U

Compliment Operator: Consider a universal set

then the compliment of the set

diagram and membership table for this operation are:

The membership table is shown below:

only when and . For other instances, Î Ç Î Î Ï Çx A B x A x B x A B

clear that the two operators union and inter-section have just contrasting characteristics

these operators are called as dual operators.

Properties with respect to Intersection Operator:

1. (idempotent law)



4. (Identity law)

A B C A B C

(Universal law)

A B A B A

Thus, we have generated a discrete structure, namely, ( ), , , ÇA B C where

, U.

: Consider a universal set U. Let A be a subset of an universal set,

the compliment of the set A is defined and described as { | }= Î ÏA x U x A

diagram and membership table for this operation are:

The membership table is shown below:

A A

1 0

0 1

A

A

Î Ç Î Î Ï Çx A B x A x B x A B . It may be

haracteristics. In

where A, B, and C are

be a subset of an universal set,

{ | }= Î ÏA x U x A . The Venn

Properties with respect to compliment Operator:

( )1. (Double negation law)

2. and

3. If , then

4. and (De-Morgan laws)

f

=

È = È = Ç = Ç =

Í Í

È = Ç Ç = È

A A

A A A A U A A A A

A B B A

A B A B A B A B

Therefore, we can claim that a collection of sets with respect to the set operators union, intersection and compliment forms a discrete structure.

Difference operator: Let A and B be two sets. Then difference of A and B is defined as

{ | and }- = Î Î ÏA B x U x A x B .

We can well claim that { | and } so that -- = Î Î Î = ÇA B x U x A x B A B A B . The Venn

diagram is shown as

The membership table can be written as

A B -A B

1 1 0

1 0 1

0 1 0

0 0 0

-A B

U

B

Symmetric Difference operator:

is defined as { | or but x }Å = Î Î Î Ï ÇA B x U x A x B A B

defined as ( ) (Å = È - Ç Å = - È -A B A B A B A B A B B A

shown as

The Membership table of ÅA B

The following are some of the properties with respect to symmetric difference operator:

1.

2. (Commutative l


fÅ =

Å = Å

Å Å = Å Å

A A

A B B A

A B C A B C

Therefore, we can say that a collection of sets with respect to symmetric difference operators forms another discrete structure.

ÇA B

Symmetric Difference operator: Let A and B be two sets. Symmetric Difference of A and B

{ | or but x }Å = Î Î Î Ï ÇA B x U x A x B A B . Equivalently,

) ( ) ( ) or Å = È - Ç Å = - È -A B A B A B A B A B B A . The Venn diagram can be

ÅA B can be written as

A B ÅA B

1 1 0

1 0 1

0 1 1

0 0 0




Therefore, we can say that a collection of sets with respect to symmetric difference operators forms another discrete structure.

U

ÅA B

Let A and B be two sets. Symmetric Difference of A and B

. Equivalently, ÅA B may be

. The Venn diagram can be


Therefore, we can say that a collection of sets with respect to symmetric difference operators

We can have some Properties with respect to Union and Intersection Operator: These

are

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

1. (Distributive law)

2.

3.

4.

5.

6.

7. ( ) (Absorption law)

8. ( )

È Ç = Ç È Ç

Ç È = È Ç È

Ç È = Ç È Ç

È Ç = È Ç È

È Ç È = Ç È Ç È Ç È Ç

Ç È Ç = È Ç È Ç È Ç È

È Ç =

Ç È =

A B C A C B C

A B C A C B C

A B C A B A C

A B C A B A C

A B C D A C A D B C B D

A B C D A C A D B C B D

A A B A

A A B (Absorption law)A

Hence, we can claim that a collection of sets to form a discrete structure with respect to the

combination of Union and Intersection Operator.

Illustrative examples:

1. Let A = {1, {1}, {2}}. Which of the following statements are true? Explain your

answer?

( ) 1 , ( ) {1} , ( ) {1} , ( ) {{1}} , ( ) {2} , ( ) {2} , ( ) {{2}} Î Î Í Í Î Í Ía A b A c A d A e A f A g A

Solution: Solution: (a) to (e) and (g) is true. The Statement (f) is false .

2 For the set A = {1, 2, 3 . . . 7}, determine the number of (i) subsets of A? (ii) proper

subsets of A? (iii) Non-empty subsets of A? (iv) Non-empty proper subsets of A (v) subsets

of A containing three elements? (vi) Subsets of A containing the elements1, 2 (vii) subsets

of A containing 5 elements including 1 and 2?

Solution: (i) Here, set A contains totally 7 elements, the experiment consists of forming

subsets using the elements of A only. Now, if we consider that B as a subset of A, then with

respect to the set B, and for an element of A, there are exactly two choices; (i) an element of A

under consideration is present in the set B or (ii) element not being present in B. Thus, each

element of A has exactly 2 choices. Therefore, total number of subsets one can construct is

72 128= . This collection includes both the null set and the set A itself (please note this).

(ii) It is known that a proper subset means it is a set a set C such that it is a subset of A and

there exists at least one element in A, not present in C. Therefore, for this reason, in the

collection of subsets, we should not include the set A. Hence, number of proper subsets of A is

72 1 127- = .

(iii) Clearly, number of non – empty subsets of A is 72 1 127- = . This is due to the fact

that we must discard the null set here.

(iv) Also, number of non – empty proper subset of A is 72 2 126- = since null set and the set

A has to be ignored here.

(v) Now, to construct all subsets of A having exactly 3 elements. This problem is equivalent

to the one, namely, in how many ways a group of 3 members may be formed from a group

containing 7 persons? The answer is given by 7 7!

353 3! 4!

æ öç ÷= =ç ÷ ×è ø

.

(vi) Here, the condition is any subset formed must include the elements 1 and 2. Therefore,

choices are there for the remaining 5 elements; either to be a part of the subset or not? Hence,

number of subsets containing 1 and 2 is 52 32= ,

(vii) Consider a five element subset of A, say B = {- , - , - , -, -}. Now, B must include the

elements 1 and 2. So, the choices of inclusion/exclusion rest with 3, 4, 5, 6, and 7 for the

remaining 3 slots. These 3 slots may be filled in 5 5!

103 2! 3!

æ öç ÷= =ç ÷ ×è ø

ways. Thus, only 10 subsets

can be formed satisfying the conditions of the problem.

3. Let S = {1, 2, 3 . . . 30}. How many subsets of A satisfy (i) |A| = 5 and the smallest

element in A is 5? (ii) |A| = 5 and the smallest element in A is less than 5?

Solution: Let B = {-, -, -, -, -} be a 5 element subset of A. As the smallest element is given

to be 5; the other 4 elements have to be greater than 5, these are to be selected from the

remaining

numbers 6 to 30 (25 in numbers). Therefore, the answer is 25 25!

12,650.4 21! 4!

æ ö÷ç ÷= =ç ÷ç ÷ç ×è ø

In the next case, the smallest element can be lower than 5 (i.e. either 1 or 2 or 3 or 4). Now if

the smallest element is 1, then the other 4 numbers may be selected in C (29, 4) ways.

If the smallest element is 2, then number of ways of selecting the other 4 numbers is C (28, 4)

ways. If the smallest element is 3, then we have C (27, 4) ways. When the smallest element is

4, then we have C(26, 4) ways. Thus, number of ways of doing the whole task is C (29, 4) + C

(28, 4) + C (27, 4) + C (26, 4).

4. How many strictly increasing sequences of integers start with 62 and end with 92?

Solution: Consider a sequence of numbers, say 1 2 3 ........... na a a a< < < <

with 1 62 and 92.na a= = The possible sequence of numbers are the following: (62, 92),

(62, 63, 92), (62, 78, 89, 92), (62, 63, 64, 65,. . 89, 90, 91, 92). In the last one all the numbers

in between 62 and 92 are included. From these, it is clear that during the construction process,

we have included some numbers in the sequence and at the same time, we have ignored the

numbers between 62 and 92. Clearly there are 27 numbers in between 62 and 92, therefore

answer to this question is 272 .

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

6, 8}, compute the following: ( ) ( ) , (b) ( ), (c) a A B C A B C C DÈ Ç È Ç È ( ) , d C DÈ

(e) ( ) , (f) ( )A B C B C DÈ - - - .

Solution: Solution: (a) Note that {1, 2, 3, 4, 5, 8}A BÈ = so that ( ) {1, 2, 3, 5}A B CÈ Ç =

.

Observe that {1, 2}B CÇ = thus, ( ) {1, 2, 3, 4, 5}A B C AÈ Ç = = . (d) By De-Morgan

law, {4, 6, 8, 9}, {1, 3, 5, 7, 9}C D= = , hence {1, 3, 4, 5, 6, 7, 8, 9}C DÈ = . (d) By

De – Morgan laws, {9}C D C DÈ = Ç = . (e) With {1, 2, 3, 4, 5, 8}A BÈ = and C = {1, 2, 3, 5, 7}, yields ( ) {4, 8}A B CÈ - = . (f) It is given that B = {1, 2, 4, 8} and C ={1, 2, 3, 5, 7}, D = {2, 4, 6, 8} gives C – D ={1, 3, 5, 7} so that B – (C -D) = {2, 4, 8}.

6. Determine the sets A and B, given that ( {1, 2, 4, 5, 7, 8, 9}A BÈ = , and

- {1, 2, 4}, - {7, 8}A B B A= =

Solution: Clearly, A = {1, 2, 4, 5, 9} and B = {5, 7, 8, 9}.

7. Determine the sets A, B and A BÈ given that {4,9}, - {1, 3, 7, 11}A B A BÇ = =

, - {2, 6, 8}B A =

Solution: {1, 3, 4, 7, 9, 11}A = {2, 4, 6, 8, 9}B = {1, 2, 3, 4, 6, 7, 8, 9, 11}A BÈ =

8. Prove or disprove the following: For the sets , , , A B C U A C B CÍ È = È implies

?A B= for sets , , , A B C U A C B CÍ Ç = Ç implies ?A B=

Solution: Consider U = {a, b, c, d, e, f, g, h} , { , }, { , , }, { , , }A d g B a e g C a d e= = = ,

observe that { , , , } and { , , , } A C a d e g B C a d e gÈ = È = but clearly A and B are different. It

may be recalled that this property holds for a set of numbers with respect to operation usual

addition, namely, a + c = b + c, then a = b. Why not here? any explanation? For the next

problem, set up U = {a, b, c, d, e, f, g, h}, { , }, { , , }, { , , }A d g B a e g C a d e= = = , note that

{ , }A C B C a eÇ = Ç = and as usual A and B are different sets. Can you explain me why these

properties are not working with sets and with respect to union or intersection operator? A

similar property is true with respect to usual multiplication operator, namely,

implying that , if 0a c b c a b c� �= = ¹ .

9. Using membership table, verify whether A B A BÈ = Ç

Solution: We shall set up :1 means , x x A BÎ and : 0 means , x x A BÏ . Consider the

membership table of A B A BÈ = Ç ,

A B A BÈ A BÈ A B A BÇ

1 1 1 0 0 0 0

1 0 1 0 0 1 0

0 1 1 0 1 0 0

0 0 0 1 1 1 1

From the above membership table (comparison of 4th column and the last column, it is clear

that A B A BÈ = Ç .

10. Using membership table, verify whether A B A BÇ = È

Solution: As above, we shall set up membership table for the set identity A B A BÇ = È by

assuming that :1 means , x x A BÎ and : 0 means , x x A BÏ .

A B A BÇ A BÇ A B A BÈ

1 1 1 0 0 0 0

1 0 0 1 0 1 1

0 1 0 1 1 0 1

0 0 0 1 1 1 1


that A B A BÇ = È .

11. Using membership table, verify whether ( ) ( ) ( )A B C A B A CÇ È = Ç È Ç .

Solution: To set up :1 means , , x x A B CÎ and : 0 means , , x x A B CÏ . Consider

A B C B CÈ ( )A B CÇ È A BÇ A CÇ ( ) ( )A B A CÇ È Ç

1 1 1 1 1 1 1 1

1 1 0 1 1 1 0 1

1 0 1 1 1 0 1 1

0 1 1 1 0 0 0 0

1 0 0 0 0 0 0 0

0 1 0 1 0 0 0 0

0 0 1 1 0 0 0 0

0 0 0 0 0 0 0 0


that ( ) ( ) ( )A B C A B A CÇ È = Ç È Ç .

12. Using Venn diagram approach, prove that

Solution: Consider

Using Venn diagram, establish that

Prof. K Gururajan, MCE, Hassan

B CÅ

z

B CÇ

B

From the above Venn diagrams, it may be noted that

13. Using Venn diagram approach, determine whether

Solution: Consider

12. Using Venn diagram approach, prove that ( ) ( ) ( )A B C A B A CÇ Å = Ç Å Ç

Using Venn diagram, establish that


AB

C

From the above Venn diagrams, it may be noted that ( ) ( ) ( )A B C A B A CÇ Å = Ç Å Ç

13. Using Venn diagram approach, determine whether ( ) ( ) ( )A B C A B A CÇ Å = Ç Å Ç

( ) ( ) ( )A B C A B A CÇ Å = Ç Å Ç

( ) ( ) ( )A B C A B A CÇ Å = Ç Å Ç

( ) ( ) ( )A B C A B A CÇ Å = Ç Å Ç


Using Venn diagram determine

whether ( ) ( ) ( )A B C A B A CÅ Ç = Å Ç Å

B CÇ

AAB

B

C C

A


( )A B CÅ ÇBA

C


A BÅ

A CÅ

C

C

A

A

B

B


( ) ( )A B A CÅ Ç Å

( )A B CÅ Ç

( ) ( ) ( )A B C A B A CÅ Ç ¹ Å Ç Å

A

A B

B

C

C

The above Venn diagrams show that ( ) ( ) ( )A B C A B A CÅ Ç ¹ Å Ç Å .

14. Use Mathematical procedure show that 1. A B A BÈ = Ç ,

Proof: To show this result, it is sufficient to prove that and A B A B A B A BÈ Í Ç Ç Í È

I.e. each one is a subset of the other. Let x A BÎ È be arbitrary. This means that

.x A BÏ È

Thus, and .x A x BÏ Ï Equivalently, and . x A x BÎ Î Therefore, x A BÎ Ç which proves

that A B A BÈ Í Ç . . . . (*1). Next consider that y A BÎ Ç be arbitrary. From this,

we obtain the results, viz., and y A y BÎ Î . Therefore, and .y A y BÏ Ï Thus,

.y A BÎ È From here, we must have A B A BÇ Í È . . . . (*2). Using (*1) and (*2), we

claim that A B A BÈ = Ç and hence the proof.

15. Give a similar proof for the following set identities:

(i) A B A BÈ = Ç (ii) A B A B- = Ç and (iii) ( ) ( ) ( ) A B C A B A CÈ Ç = È Ç È

A Discussion on Duality Concept:

Consider a set expression, say, X, involving the operators such as union, intersection,

compliment etc. The dual of X may be obtained by just replacing union by intersection and

intersection by union without making any changes in the others. However, if an expression

contains a special symbol like U (universal set), this is to be replaced by the null set f . For

example, the dual of ( )A A BÇ Ç is ( )A A BÈ È . The dual of ( ) ( )X X Y A BÈ Ç È Ç is

( ) ( )X X Y A BÇ È Ç È . Dual of ( ) ( )A A B A f= È Ç È is ( ) ( )A A B A U= Ç È Ç .

15. Using laws of set theory, simplify the expression ( ) ( )( )A B A B C D A BÇ È Ç Ç Ç È Ç .

Solution: Here, we shall use absorption law of set theory, namely, ( )X X Y XÈ Ç = . First, we

shall set up , ,X A B Y C D= Ç = Ç then above can be modified as ( ) ( )X X Y A BÈ Ç È Ç .

Now applying absorption law, we obtain ( )X A BÈ Ç . Now substituting for X, yields,

( ).A B A BÇ È Ç . Now expanding this using distribution law, we get =

( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( )( ) ( )

because

by Associative law

A B A B A A A B B A B B

U A B B A B

A B B A B U X X

A B B B A

Ç È Ç = È Ç È Ç È Ç È

= Ç È Ç È Ç

= È Ç È Ç Ç =

é ù= È Ç Ç Èë û

= ( ) by Absorption law

by Absorption law

B B A

B

Ç È

=

16. ( ) ( )Simplify A B A B CÈ È Ç Ç


| | | | | | | |A B A B A BÈ = + - Ç

A BA BÇ

Inclusion – Exclusion Principle

Note that while counting the elements of the set A BÈ , number of elements in A BÇ is

counted twice; once for while counting the elements of the set A and next time for the set B. In

view of this, we get | | | | | | | |A B A B A BÈ = + - Ç . This is called Inclusion – Exclusion

Principle for two finite sets. For 3 sets A, B, and C, the inclusion – exclusion principle may be

stated as

| | | | | |A B C A B C A B A C B C A B CIÈ È = + + - Ç - - Ç + Ç Ç . The proof can be

given as follows: We shall set up A B DÈ = . Using this above, we get

( )

( )

( ) ( )

| |

=

(using distributive law here)

A B C D C

D C D C

A B C A B C

A B A B C A B C

A B A B C A C B C

È È = È

+ - Ç

= È + - È Ç

= + - Ç + - È Ç

= + - Ç + - Ç È Ç

( ) ( )

( ) ( )

(But ), thefore, we obtain

A B A B C A C B C A C B C

A C B C A B C

= + - Ç + - Ç + Ç - Ç Ç Ç

Ç Ç Ç = Ç Ç

( )A B C A B C A B A C B C A B CÈ È = + + - Ç - Ç - Ç + Ç Ç

This is the Inclusion – Exclusion Principle for 3 sets.

Illustrative Examples:

A computer company must hire 25 programmers to handle system programming jobs

and 40 programmers for applications programming. Of those hired, ten will be expected

to perform jobs of both types. How many programmers must be hired?

Solution: Solution: Let A and B denote the number of programmers handling system

programming and applications prog

25, 40 and 10.A B A B= = Ç =

programmers required to handle either of these two jobs i.e. to find

Inclusion – exclusion principle for 2 sets, viz.,

| | 25 40 10 55A B A B A BÈ = + - Ç = + - =

hired. The same situation can well be explained by considering the following Venn diagram;

From this Venn diagram, it is

programming jobs is 15, while the number of programmers knowing application programming

is 30. Also, number of programmers capable of handling exactly one of the two jobs is 45. On

the other hand, suppose company hires say 60 programmers, then it follows that 5

programmers do not have any knowledge on either of the types of jobs.

In a survey of 260 college students, the following data were obtained: 64 had taken a

Mathematics course, 94 had taken a C

course; 28 had taken both Mathematics and a Business course. 26 had taken a

Exclusion Principle for 3 sets.


mers for applications programming. Of those hired, ten will be expected

to perform jobs of both types. How many programmers must be hired?

Solution: Let A and B denote the number of programmers handling system

programming and applications programming respectively. According to the given data,

25, 40 and 10.A B A B= = Ç = Here, the problem is to compute the number of

programmers required to handle either of these two jobs i.e. to find A BÈ

exclusion principle for 2 sets, viz.,

| | 25 40 10 55È = + - Ç = + - = . Thus, totally 55 programmers are to be


From this Venn diagram, it is clear that number of programmers who knows the system



ppose company hires say 60 programmers, then it follows that 5

programmers do not have any knowledge on either of the types of jobs.


Mathematics course, 94 had taken a Computer science course; 58 had taken a Business



mers for applications programming. Of those hired, ten will be expected

Solution: Let A and B denote the number of programmers handling system

ramming respectively. According to the given data,

Here, the problem is to compute the number of

.A B Consider the

exclusion principle for 2 sets, viz.,

Thus, totally 55 programmers are to be


clear that number of programmers who knows the system



ppose company hires say 60 programmers, then it follows that 5


omputer science course; 58 had taken a Business


Mathematics and Computer science course, 22 had taken both a Computer science and a

Business course, and 14 had taken all three types of courses. Then how many students

were surveyed who taken none of the three types of courses? How many had taken only a

computer science course?

Solution: Let M, C and B denote the number of students who have registered for Mathematics,

Computer Science and Business course respectively. From the given data, it may be noted

that

| | 64, | | 94, | | 58, | | 28, | | 26, | | 22 and | | 14M C B M B C M C B M C B= = = Ç = Ç = Ç = Ç Ç =

Using Inclusion – Exclusion principle here,

| | | | | | | | | | | | | | | | 154M C B M C B M C M B C B M C BIÈ È = + + - Ç - - Ç + Ç Ç = . Thus,

154 students have registered for either one of the three courses. As the number of students in

the college is 260, number of students who have not registered for any one of these courses is

106. Number of students studying only computer science is 60. Also, it may be noted that

number of students studying exactly two of the three subjects is 34. Number of students

studying exactly one of the three subjects is 106.


M C

B

24 121414 8

60

260N =

M: Mathematics

C: ComputerScience

B: Business course

| | 64, | | 94,M C= =| | 58, | | 28B M B= Ç =

| | 26, | | 22C M C BÇ = Ç =| | 14M C BÇ Ç =

22

30 cars were assembled in a factory. The options available were a radio, an air

conditioner, and white-wall tires. It is known that 15 of the cars have radios, 8 of them

have air conditioners, and 6 of them have white-wall tires. Moreover, 3 of them have all

three options. How many cars do not have any options at all?

Solution: Let A, B, and C denotes the number of radios, air conditioners and white – wall tires

available for a car in a factory, respectively. From the given data, we have N = 30,

| | 15, | | 8, | | 6 and | | 3A B C A B C= = = Ç Ç = . First we shall to do some groundwork to

solve this problem. , so that ,A B C A B A B C A BÇ Ç Í Ç Ç Ç £ Ç

, so that ,A B C A C A B C A CÇ Ç Í Ç Ç Ç £ Ç

, so that ,A B C B C A B C B CÇ Ç Í Ç Ç Ç £ Ç 3 , 3 , 3 , A B A C B C£ Ç £ Ç £ Ç

Thus, 9 (multiply throughout by -1)A B A C B C£ Ç + Ç + Ç yields,

9A B A C B C- Ç - Ç - Ç £ - . Adding | | | | | | | | on both sides,A B C A B C+ + + Ç Ç

| | | | | | | | | | | | | | 9 | | | | | | | | A B C A B A C B C A B C A B C A B C+ + - Ç - Ç - Ç + Ç Ç £ - + + + + Ç Ç

i.e. | | 9 15 8 6 3 23. A B CÈ È £ - + + + + = Equivalently, at most 23 cars have one of these

options. Therefore, 7 cars will not have any options at all.

A Mathematics Professor gave a class test consisting of three questions. (I, II and III).

There are 21 students in his class, and every student answered at least one question. Five

students did not answer the first question, seven failed to answer the second question, and

six did not answer the third question. If nine students answered all three questions, then

how many answered exactly one question?

Solution: Let a, b, c denote respectively the number of students who have answered exactly 1

of the 3 questions. Let d, e, f denote the number of students who have answered exactly 2 of

the 3 questions. It is given that 9 students have answered all the 3 questions. These situations

may well be explained by means of a Venn diagram shown below.


I II

III

ab

c

d

ef 9

a + b + c + d + e + f + 9 = 21 or a + b + c + d + e + f = 12 (1) b + c + f = 5

(2) a+ c + e = 7 (3) a + b + d = 6 (4) Now consider (2) + (3) + (4). This yield, 2 (a

+ b + c) + d + e + f = 18. The same can be written as (a + b + c + d + e + f) + (a +b + c) = 18.

Using (1) here, we obtain a + b + c = 6. Thus, it is clear that only 6 students have answered

exactly one of the three questions.

A discussion of Discrete Probability

Deterministic approach: Given a problem, one can consider two ways of solving it; (i) using

Deterministic procedure and ii) by using Probabilistic Approach. If information /data /

parameters of a problem is known in advance, the method that is used to solve it is called as

deterministic approach. For example, the previous problem discussed above, where we have

solved the problem using a deterministic approach.

When we do not have complete information about a problem, and if we try to find a solution to

it, the procedure that is adopted is referred to as probabilistic approach. For example, consider

the experiment of finding how much this programme will be beneficial to students? What are

their chances of doing well in the exams scheduled in December– 2012?

I would say that only problems of this kind have received a considerable attention from

everyone including researchers and engineers. In fact, people say in a number of platforms that

“Past is history, Present is Manageable but Future is interesting or un–certain”.

A discussion of Discrete Probability

Sample space: The set of all outcomes of a random experiment is called sample space,

denoted by S.

An event in a sample space: Any subset or a part of a sample space is called an event.

.e. if , then is called an event in the sample space.A S AÍ

Probability function associated with an event: With every event, A in a sample space, S, we

assign a real number in between 0 and 1, called the probability/chance of occurrence of the

event A. This is denoted by ( )P A .

Thus, ( )P A is a function from S to [0, 1]. This function ( )P A has the following properties

called as probability axioms:

• For any event, A, ( ) 0P A ³

• ( ) 1P S =

• If A and B are any two mutually exclusive events in S i.e. A B fÇ = , then

( ) ( ) ( )P A B P A P BÈ = +

• In general, if 1 2 3, , . . . kA A A A are any collection of k mutually exclusive events

then ( ) ( ) ( ) ( )1 2 3 1 2 3( . . . ) . . . k kP A A A A P A P A P A P AÈ È È È = + + + +

• On the other hand, if A and B are not mutually exclusive events, then

( ) ( ) ( ) - ( )P A B P A P B P A BÈ = + Ç

The chance of occurrence of an event, say, A in the sample space is defined as

number of ways in which event occursnumber of ways in which sample space, occurs

|)

|| |

( A

SA

AP

S= = .

Before proceeding to further discussion on probability, let us have a discussion on some

important counting principles:

1. Rule of sum: Let T1 and T2 be different two tasks such that T1 can be done in n1 ways and

T2 in n2 ways (say), then either T1 or T2 can be performed in exactly n1 + n2 ways.

Example: Suppose that a computer show room has 25 Laptops produced by Dell and there

are 30 Laptops produced by Asus. A customer visits the show room and he is interested to buy

a laptop. How many choices are there for him?

Since, customer can buy a laptop produced by either one of the companies, clearly customer

has 30 + 25 = 55 choices.

Rule of Multiplication: Now consider the same problem but with a difference. Suppose that

customer wants to buy 1 laptop produced by Dell and 1 laptop produced by Asus. As there are

30 Asus laptops and 25 Dell laptops, the customer has 30 25 750× = choices. This is called

multiplication rule.

Rule of permutation: Consider a set A with ndifferent elements. Then these n elements may

be arranged in ! 1 2 3 . . . n n= × × ways. A selection of any (0 )r r n< < elements from A and

then arranging these in some order is called r – permutation of A. It is denoted by

! or ( , )

( )!r

nnP P n r

n r=

-.

Rule of combination: Consider a set A with n elements. A mere selection of any

(0 )r r n< < elements from A is called r – combination of A. It is denoted by

( )n !

or ( , ) or r ! !r

nnC C n r

n r r

æ ö÷ç ÷=ç ÷ç ÷ç - ×è ø.

Generalized Permutation Principle: Consider a set A with n objects,

1 2 3{ , , . . . }n

A x x x x= such that 1n objects are alike i.e. having same

behaviors, 2n objects are alike i.e. having similar behaviors, 3n elements are alike and so on

such that 1 2 3 . . . . n n n nk n+ + + + = , then, all the elements of A

can be arranged in 1 2 3

!! ! ! . . . !k

n

n n n n× × × ways.

Example: Find the number of ways arranging the letters of the word MISSISSIPPI.

Here, note that letter M occurs once, I 4 times, S occurs 4 times, P occurs twice. In view of

these, the letters of the word MISSISSIPPI can be arranged in 11!

4! 4! 2!1!× × × ways.

If the letters of the word BOOLEAN are arranged at random, what is the probability

that the two O’s remain together in the arrangement?

Solution: Here, experiment consists of arranging the letters of the word, B, O, O, L, E, A, N.

Let S denote the sample space of the experiment, then as O appears twice, other letters

appearing only once, clearly 7!

2520.2!

S = = Let A denote the event, namely, that, two O’s

always appear together in the arrangement. Consider two O’s as one block, thus, we have 6

different types of letters in the arrangement which can be shown as

OO B L E A N

Thus, these six letters may be arranged in 6! = 720 ways. Therefore, the probability of the

event may be calculated as A

P AS

= = = =| | 720 2

( ) 0.2857| | 2520 7

.

If the letters of the word in the acronym WYSIWYG (What You See Is What You Get)

are arranged in a random manner, then what is the probability that starts and ends with

the same letter?

Solution: Here, too experiment consists of arranging the letters of the word W, Y, S, I, W, Y,

G. Let S denotes the sample space of the experiment. Observe that the letters W and Y occurs

twice, and other letters occurs only once, therefore, S = =×7!

| | 12602! 2!

. Form the event A:

“An arrangement of letters W, Y, S, I, G starts and ends with the same letter” Possible

arrangement could be either

W --- --- --- --- --- W

Y --- --- --- --- --- Y

Clearly, |A| = + =5! 5!

1202! 2!

, therefore, A

P AS

= = =| | 120

( ) 0.0952| | 1260

.

In a survey of 120 passengers, an airline found that 48 enjoyed wine with their meals, 78

enjoyed mixed drinks, and 66 enjoyed iced tea. In addition, 36 enjoyed any given pair of

these beverages and 24 passengers enjoyed them all. If 3 passengers are surveyed at

random? What is the probability that All 3 enjoy iced tea with their meals? (ii) All 3

enjoy only iced tea with their meals? (iii) How many passengers enjoy exactly one of the

three beverages served?

Solution: Let W, M, I denote the set of Airline Passengers who enjoy Wine, Mixed Drinks and

Iced Tea with their meals respectively. From the data, we have W M I= = =48, 78, 66,

W M W I M IÇ = Ç = Ç = 36,


120N =

W M

I

24

12

1212

0 30

18

W: Wine, M: Mixed Drinks I: Ice Tea

48, 78, 66,W M I= = =

36

36

36

W M

W I

M I

Ç =

Ç =

Ç =

24 W M IÇ =I

(a) Note that 36 passengers enjoy exactly two of the three beverages along with their

meals.

(b) Note that 48 passengers enjoy exactly one of the three beverages with their meals.

(c) Note that only 30 passengers enjoy Mixed drinks with their meals.

(d) Note that 108 passengers enjoy either one of the three beverages served with their

meals.

Let S be the sample space of the experiment, then S C= = =×

120!(120, 3) 280840

117! 3!. Let

A: “Selected all 3 passengers enjoy iced tea with their meals”. Then

A C= = =×

66!(66, 3) 45760

63! 3!. Therefore,

AP A

S= = =

45760( ) 0.1630

280840. Let B: “All

3 selected passengers chosen at random enjoy only iced tea with their meals”. From the

given data, it is clear that only 18 passengers enjoy iced tea with their meals, therefore,

B C= = =×

18!(18, 3) 816

15! 3!. Therefore,

BP B

S= = =

816( ) 0.03915

280840.

Problem: If 3 integers are selected at random from S = {1, 2, 3. . . , 99, 100}, what is

the probability that their sum is an even integer?

Solution: Let x y z, , be three integers selected at random from S consisting of first 100

positive integers. Let S be the sample space of the experiment. Then, cardinal number of S

can be S C= = =×

100!(100, 3) 161700

97! 3!.

Set up A: x y z+ + is an even integer. We know that sum of three integers can be even

only when all the integers are even or one integer is even, while the other two are odd.

Keeping this in view, we can consider when (i) x y z, , all are even integers and (ii) when

one integer is even and the other two are odd integers. For case (i), three even integers

can be chosen from the available 50 even integers in C = =×

50!(50, 3) 19600

47! 3! ways. For

case (ii) 1 even integer can be selected in 50 ways and as there are 50 odd integers, 2 odd

Integers can be chosen in C = =×

50!(50, 2) 1225

48! 2! ways. Therefore, number of ways

making selection such that one integer is even and other two are odd is ×50 1225=61250 .

Therefore, A = + =61250 19600 80850 . Hence, A

P AS

= = =80850

( ) 0.50161700

.

The freshman class of a private engineering college has 300 students. It is known that

180 can program in Pascal, 120 in FORTRAN, 30 in C++, 12 in Pascal and C++, 18 in

FORTRAN and C++, 12 in Pascal and FORTRAN, and 6 in all three languages. (a) A

student is selected at random. What is the probability that she can program in

exactly two languages? (b) Two students are selected at random. What is the

probability that they can (i) both can program in Pascal? (ii) Both program only in

Pascal?

Solution: Let P, F, C denote the set of students who can program Pascal, FORTRAN and

C++ respectively. From the data, we have

P F C P F P C C F= = = Ç = Ç = Ç =180, 120, 30, 12, 12, 18,


180, 120, 30,P F C= = =

12

12

18

P F

P C

F C

Ç =

Ç =

Ç =

6 P F CÇ =I

300N =P F

C

6

6

6 12

162 96

6

The experiment consist of selecting one student at random from the college strength of 300

and then testing his programming skills in 3 types of languages, namely, Pascal,

FORTRAN, C++. Let S be the sample space of the experiment, then clearly S = 300 . Set

up A: The selected student can program in exactly two of the three languages. Then,

clearly, A = 24 . Thus, the occurrence of the event A is given as P A = =24

( ) 0.08300

.

(b) Here, the experiment consists of selecting two students from the college of strength

300, and then finding their skills with regard to programming. Let S be the sample space of

the experiment. Then, ( )S C×

= = = =×

300! 299 300300, 2 44850

298! 2! 2. Set up B: “the

selected two students can do programming in Pascal”, then

( )B C×

= = = =×

180! 179 180180, 2 16110

178! 2! 2, therefore,

BP B

S= = =

16110( ) 0.3592

44850.

(c) Set up C:”The chosen two students can do programming only in Pascal”, then

clearly ( )C C×

= = = =×

162! 161 162162, 2 13041

160! 2! 2. Thus,

CP C

S= = =

13041( ) 0.2908

44850

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