The objectives of Discrete Mathematical Structures are · The objectives of Discrete Mathematical...
Transcript of The objectives of Discrete Mathematical Structures are · The objectives of Discrete Mathematical...
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.
The functioning of union operator may also be explained using membership table and Venn
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
The functioning of union operator may also be explained using membership table and Venn
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)
3. ( ) ( ) (Associative 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
same can be explained by using Venn diagram and by employing a membership table.
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
same can be explained by using Venn diagram and by employing a membership table.
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
3. ( ) ( ) (Associative law)
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)
2. (Commutative law)
3. ( ) ( ) (Associative 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
3. ( ) ( ) (Associative law)
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
The following are some of the properties with respect to symmetric difference operator:
2. (Commutative law)
3. ( ) ( ) (Associative law)
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
The following are some of the properties with respect to symmetric difference operator:
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
From the above membership table (comparison of 4th column and the last column, it is clear
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
From the above membership table (comparison of 5th column and the last column, it is clear
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
Prof. K Gururajan, MCE, Hassan
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
Prof. K Gururajan, MCE, Hassan
Using Venn diagram determine
whether ( ) ( ) ( )A B C A B A CÅ Ç = Å Ç Å
B CÇ
AAB
B
C C
A
Prof. K Gururajan, MCE, Hassan
( )A B CÅ ÇBA
C
Prof. K Gururajan, MCE, Hassan
A BÅ
A CÅ
C
C
A
A
B
B
Prof. K Gururajan, MCE, Hassan
( ) ( )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
Prof. K Gururajan, MCE, Hassan
| | | | | | | |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.
A computer company must hire 25 programmers to handle system programming jobs
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
hired. The same situation can well be explained by considering the following Venn diagram;
From this Venn diagram, it is clear that number of programmers who knows the system
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
ppose 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 Computer science course; 58 had taken a Business
course; 28 had taken both Mathematics and a Business course. 26 had taken a
A computer company must hire 25 programmers to handle system programming jobs
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
hired. The same situation can well be explained by considering the following Venn diagram;
clear that number of programmers who knows the system
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
ppose company hires say 60 programmers, then it follows that 5
In a survey of 260 college students, the following data were obtained: 64 had taken a
omputer science course; 58 had taken a Business
course; 28 had taken both Mathematics and a Business course. 26 had taken a
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.
Prof. K Gururajan, MCE, Hassan
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.
Prof. K Gururajan, MCE, Hassan
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,
Prof. K Gururajan, MCE, Hassan
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,
Prof. K Gururajan, MCE, Hassan
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