AAU

2014

Inclusion Exclusion principle

Getenet Dessie

MiliyonNew Stamp

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ACKNOWLEDGEMENT

Firstly, I would like to express my heartful gratitude towards my Advisor

Ato Biadgligne Asmara for his continuous advice in preparing this project.

Secondly, I would like to thank my families and all my friends for their

Indispensible financial and moral support.

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THE OBJECTIVE OF THIS PROJECT

a. To exactly count the number of elements in the union of finite number of sets where the sets are disjoint sets (or where the sets do not have any element in common).

b. To correctly count the number of elements in the union of finite number of sets where the sets are not disjoint (or when there is a common element between sets).

c. To derive a formula for calculating the number of elements in the union of finite number of sets where the sets are whether disjoint or not.

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CONTENTS

Title Page

CHAPTER ONE:

Introduction:..5 Basic counting principle:5 The product rule:.5 The sum rule:.6

CHAPTER TWO:

2. The inclusion and exclusion:7

2.1 Number of elements for two non disjoint sets:8

2.2 Proof for the number of elements in the union of three sets:.12

2.3 Theorem (the inclusion and exclusion principle):..14

2.4 Proof of the inclusion and exclusion principle:.15

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CHAPTER ONE

INTRODUCTION

Combinatorics is the branch of discrete mathematics that explores counting, combining and arranging the elements of sets. Enumeration may be thought of as counting but in a broader sense it is often included with combination and permutation as one aspect of Combinatorics, with graph theory being the other aspect. In Combinatorics counting may be a simple process of adding the number of elements of two disjoint sets, or it may be a more complex process, as we will see in the principle of Inclusion and Exclusion.

Combinations are collections of new sets at elements. Permutations are the arrangement of elements within sets. The field of Combinatorics has practical applications in computer science, probability

. Basic counting principle

We will present two basic counting principles, the product rule and the sum rule. Then we will show how they can be used to solve many different counting problems problems.

The product rule applies when a procedure is made up of separate tasks.

The product rule

Suppose that a procedure can be broken down in to a sequence of two tasks. If there are ways to do the first task and for each of these ways of doing the first task, there are ways to do the second task, then there are ways to do the procedure.

Example: A new company with just two employees, Sanchez and Patel rents a floor of a building with 12 offices. How many ways are there to assign different offices to these two employees?

Solution: the procedure of assigning offices to these two employees consists of assigning an office to Sanchez, which can be done in 12 ways, then assigning an office to Patel different from the office assigned to Sanchez, which can be

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done in 11 ways. By the product rule, there are 12.11=132 ways to assign offices to these two employees.

Example: the chairs of an auditorium are to be labeled with a letter and a positive integer not exceeding 100. What is the largest number of chairs that can be labeled differently?

Solution: the procedure of labeling a chair consists of two tasks, namely, assigning one of the 26 letters and then assigning one of the 100 possible integers to the seat. The product rule shows that there are 26.100=2600 different ways that a chair can be labeled.Therefor; the largest number of chairs that can be labeled differently is 2600.

THE SUM RULE

If a task can be done either in ways or in one of ways ,where non of the set of ways is the same as any of the set of ways, then there are + ways to do the task.

Example: suppose that either a member of the mathematics department or a student who is mathematics major is chosen as a representative to a university committee. How many different choices are there for this representative of there are 37 members of the mathematics department and 83 mathematics majors and no one is both a member and a student?

Solution: there are 37 ways to choose a member of the mathematics department and there are 83 ways to choose a student who is mathematics major. Choosing a member of the mathematics department is never the same as choosing a student who is mathematics major because no one is both a department member and a student. By the sum rule, it follows that there are 37+83=120possible ways to pick this representative.

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CHAPTER TWO

THE INCLUSION AND EXCLUSION PRINCIPLE

Suppose that a task can be done in or in ways but that some the set of ways to do the task are the same as some of the other ways to do the task. In this situation, we cannot use the sum rule to count the number of ways to do the task. Adding the number of ways to do the tasks in these two ways leads to an over count, because the ways to do the task in the ways that are common are counted twice. To correctly count the number of ways to do the two tasks, we add the number of ways to do it in one way and the number of ways to do it the other way, and then subtract the number of easy to do the task in a way that is both among the set of ways and the set of ways.This technique is called the inclusion and exclusion principle.

The principle of inclusion and exclusion is a way of thinking about combining sets with overlapping elements. First let as consider two sets, the set of girls in a class room and the set of boys in a class room .These are disjoint sets. They share no elements in common; a student could not be a member of both sets. The Addition principle states that if two sets are disjoint, the size of the union of both sets equals the number of elements in the second set. This can be stated as

A =A+B .

In our class room example we will assume that A contains 10 girls and the set B contains 12 boys.

So, = + = 10+ 12 = 22

There are 22 items in the set .

Now let us consider two different class room sets. The set of challenge students and the set of band students. Let A be the set of band students; there are 8 students in this set. Let B be the set of challenge students ,and there are 6 students in this set. Can we combine the two sets and use the Addition principle to count 14 students in the set of in the of ? In this example

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,the sets are not disjoint . It is quite possible for a challenge student to also be a band student .We cannot use the Addition principle , but the principle of Inclusions and Exclusion gives us a strategy .Let us follow a visual representation to our solution.

It will count all the members of set A and combine that with all the members of set B.In combining these two sets,we are obviously over counting the total ,since two students belonge to both sets.To compensate ,we will subtract the number of students who are in both set A and set B.We starting by including all elements ,then exclude elements common to both sets .Putting numbers to the word we have 8+6-2=12 band and challenge students.

Conceptually we have that the number of elements in the union of the two sets A and B is the sum of the numbers of elements in the sets minus the number of elements in their intersection.That is

= +

The formula for the number of elements in the union of two sets is useful in counting problems.

Example: In a discrete mathematics class every student is a major in computer science or mathematics or both.The number of students having computer science as a major (possibly along with mathematics ) is 25.The number of students having mathematics as a major (possibly along with computer science)is 13 ;and the number of students majoring in both computer science and mathematics is 8.How many students are in this class?

Solution: let A be the set of students in the class majoring in computer science and B be the set of students in the class majoring in mathematics. Then is the set of students in the class who are joint mathematics and computer science majors. Because every student in the class is majoring in either computer science or mathematics(or both)it follows that the number of students in the class is .

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There for,

= +

25+13-8=30

Therefor,there are 30 students in the class.This computation is illustrated in figure 1:

= +

=25+13-8=30

AB

||=25 |A B| =8 ||=1

Figure1:The set of students in a discrete mathematics class.

Example: how many positive integers not exceeding 1000 are divisible by 7 or 11?

Solution: Let A be the set of positive integers not exceeding 1000 that are divisible 7,and B be the set of positive integers not exceeding 1000 that are divisible by 11.Then is the set of integers not exceeding 1000 that are divisible by either 7 or 11 and is the set of integers not exceeding 1000 hat are divisible by both 7 d 11.Among the positive integers not exceeding

1000 there are

integers divisible by 7 and

divisible by 11.Because

7and 11 are relatively prime ,the integers divisible by both 7 and 11 are those

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divisible by both 7.11.Consequently,there are

. positive integers not

exceeding 1000 that are divisible by both 7 and11.It follows that there are

= +

=

+

+

.

=142+90-1

=220

Positive integers not exceeding 1000 that are divisible by either 7 or 11.This computation is illustrated in figure 2.

= +

=142+90-12

=220

||= 142 | |= 12 ||= 90

FIGURE2: The set of positive integers not exceeding 1000 that are divisible by either 7 or 11.

Example3:shows how to find the number of elements in a finite universal set that are out side the union of sets.

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Example3:suppose that there are 1807 freshmen at your school .Of these 453 are taking a course in computer science, 567are taking a course in mathematics, and 299 are taking a course in both computer science and mathematics. How many are not taking a course either in computer science or in mathematics?

Solution: To find the number of freshmen who are not taking a course in either mathematics or computer science , subtract the number that are taking a course in either of these subjects from the total number of freshmen.

Let A be the set of all freshmen taking a course in computer science and let B be the set of all freshmen taking a course in mathematics.

It follows that |A|=453,|B|=567,| |=299.

The number of freshmen taking a course in either computer science or mathematics is :

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

=453+567-299

=721

Consequently, there are 1807-721=1086 freshmen who are not taking a course in computer science or mathematics.

We will now begin our development of a formula for the number of elements in the union of a finite number of sets. The formula we will develop is called the principle of inclusion and exclusion .For concreteness, before we consider union of n sets, where n is any positive integer ,we will derive a formula for the number of elements in the union of three sets, A ,B ,and C.

To construct this formula we note that |A|+|B|+|C| counts each element that is in exactly one of the three sets once,elements that are in exactly two of the sets twice ,and elements in all three sets three times. This is illustrated in the first panel in figure 3.

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a)count of elements b)count of elements c)count of elements by By |A|+|B|+|C|

To remove the over count of elements in more than one of the sets we subtract the number of elements in the intersections of all pairs of the three sets .We obtain

+ +

This expression still counts elements that occur in exactly one of the sets once. An element that occur in exactly two of the sets is also counted exactly once ,because this element will occur in one of the three intersections of sets taken two at a time. However, those elements that occur in all three sets will be counted zero times by this expression, because they occur in all three intersections of sets taken two at a time. This is illustrated in second panel in figure 3.

To remedy this under count, we add the number of elements in the intersections of all three sets. This final expression counts each element once, whether it is in one, two, or thereof the sets .Thus,

+ + +

This formula is illustrated in the third panel of figure three

Now let us expand our example to three sets that are not disjoint ;band students, challenge students and strings students.

By |A|+|B|+|C|-| |-| |-| |

By |A|+|B|+|C|-| |-| |-| |+| |

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Let A be the set of band students, B be the set of challenge students and C is the set of strings students.

We can start by adding the members of all the three sets.

+ + = 8 + 6 + 5 = 19

In counting all members of the three sets we have obviously over counted again.

We will compensate by subtracting out, or excluding, the members that are in both band and challenge or = 2. likewise, we will exclude the common members of band and strings, = 2, and the common members of challenge and strings , = 3.

Consider the overachiever in band, challenges and strings. We have added him/her in three times as a member of the individual sets, but we have also subtracted him/her out three times as a common member of all three groups. We need to this student in, or = 1. We have

8 + 6 + 5 2 2 3 + 1 = 13 Total students.

Stringing together our process we have

= + + +

Example :A total of 1232 students have taken a course in Spanish, 879 have taken a course in French, and 114 have taken a course in Russian. Further,103 have taken a course in both Spanish and French ,23 have taken a course in both Spanish and Russian ,and 14 have taken a course in both French and Russian. If 2092 students have taken at least one of Spanish ,and Russian ,how many

students have taken a course in all three languages?

Solution: let S be the set of students who have taken a course in Spanish, F the set of students who have taken a course in French and R the set of students who have taken a course in Russia. Then

S=1232, F=879, R =114

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SF=103, SR=23, FR=14, and SFR=2092 When we insert this quantities in to the equation.

SFR=S+F+R-SF-SR-FR+-SFR

|F| = 879

| |= 103

||= 1232

| |=23 | |= 114

||= 114 | | = 2092

| |=?

fig 4:- The set of students who have taken courses in Spanish , French , Russian

2092=1232+879+114-23-103-14+SFR

we now solve for SFR. We ind that SFR=7. Therefore, there are 7 students who have taken courses in Spanish, French and Russian.

we will now state and prove the inclusion and exclusion principle which tells us how many elements are there in the union of finite number of finite sets.

Theorem (The Inclusion-Exclusion principle) :-Given n = the number of finite sets Ai where 1 n ,we can generally state the principle of Inclusion and Exclusion

as

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= +

.+(1)

As the principle of inclusion and exclusions is extended to a greater number of sets ,it remains a process of including everything in all sets ,then excluding the overlap of items common to sets.

A five set example would include all the individual items in five sets, exclude the intersection of each pair of sets, include the intersection of each triple of sets, exclude the intersection of each of each quadruple of sets ,and finally include the intersection of the quintuple of sets.

The principle of Inclusion and Exclusion can be proved through the use of combinatorial approach

Let us consider X,an element of set S.X belongs to r subset of S,which we will call . We do not need to consider any that does not X, as they do not contribute to the union of sets we are attampting to count .

We want to count X once on the left side of the equation in the union of sets ,so we need to count X one time on the right side of the equation ,since X is in r subsets, the first term of the right hand side of the equation is r.The second term subtracts all of X in the intersection of all paired sets with in r, the third term adds back in occurrences of X in the intersection of three sets, the fourth term subtracts occurrences of X in four set intersections.

We can show the number of occurrences for each term as C(r,k) and state this as:

1=C(r,1)-C(r,2)+C(r,3)-...(1)(,)

1-C(r,1)+C(r,2)-C(r,3+...+(1)(,)= (1 1) = 0

The binomial theorem showing that both sides of the equation are the same. This confirms we are adding X appropriately when using the principle of Inclusion and Exclusion.

Let's apply the principle of Inclusion and Exclusion in same typical situation

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Our first investigation concerns prime factors

Given that the set of positive integer from 1 to 1000 how many integers are evenly divisible by a single digit prime number?

To evaluate the size of sets A,B,C and D ,we need to count the factors when 1000 is divided by the prime numbers in question. If we divide 1000 by our number and truncate any remainder ,we have the quantity we are looking for. There is an appropriate notation for this, the greatest integer or floor, function , which return the greatest integer less than the value inside the half brackets.

If we wanted to know how many times 3 will divide in to 10, then 10 3 =3 is

how we can calculate and show the result.

The principle of inclusion and exclusion gives us the strategy to solve our problem .We will start by including all members divisible by 2, 3, 5 or 7.But in doing so we will over count.

For example, the number 30 will appear in the set of numbers divisible by 2, set of numbers divisible by 3, and the set of numbers divisible by 5.We will compensate by subtracting the elements appearing in all two set intersections, then we will add back in the elements that are member of three set intersections, and finally subtract any four set intersections .We can express this as:

Let N={1,2,3, 1000}

Let A={ N X is divisible by 2}

Let B={ N X is divisible by 3 }

Let C={ N X is divisible by 5}

Let D={ N X is divisible by 7}

=A+B+C+D- - - - -

- + + + + -

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We will use the process noted above to determine the size of the sets .For the size of the set intersections, we will consider that an integer is divisible by both a and b when and only when it is divisible by the LCM (least common multiple) of a and b.

=

=500 =

=333, =

=200

=

=142 =

=166 =

=100

=

=71 =

=66 =

=47

=

=28 =

=33 =

=23

=

=14 =

=9

=

=4

Placing our values in to our formula we have

= 500+ 333+ 200+ 142 166 100 71 66 47 28+ 33+ 23+ 14+ 9 4 = 772

So, 772 integers from the set N are divisible by a single digit prime numbers.

Our next example uses the principle of inclusion and exclusion in a similar way.

Consider a trio of golden summer evenings at the college world series.The often dance gate for the three championship college world series game was 75,012.Aurvey showed that 60,834 individuals attend at least one championship game and 2,578 attend all three. How many fans attend exactly two games of the championship?

Let A be the set of people who attend game 1.

Let B be the set of people who attend game 2.

Let C be the set of people who attend game 3.

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The principle of inclusion and exclusion states the union of these three sets , that is the number of people who attend at least one game of the series is equivalent to the sum of A,B,Cless the sum of the individuals who attend two games plus the people who attend all three games . So

=A+B+C- - +

=60,834 fans attended at least one championship game .

A+B+C=the gate for three games=75,012 fans .

=2,578 fans attended all three games .So,

60,834=75,012-X+2,578 and

X=16,756

So, although we can not say which two games they attended, we do know that 16,756 people attended at least two college world championship games. Of these 16,756 people who attended at least two games ,2578 attended all three.So,14,178 fans attended exactly two games.

Example:-how many elements are there in the union of four sets if the sets have 50,60,70, and 80 elements respectively, each pair of the sets has 5 elements in common ,each triple of the sets has 1,common elements ,and no element is in all four sets ?

Solution Let A be the first set

Let B be the second set

Let C be the third set

Let D be the fourth set

A=50 , B=60 , C=70, D=80

A =A =A =B =B =C =5

A =A =A = =1

A =0

using the Inclusion and exclusion principle

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=A+B+C+ D-

A -A -A -B -B -C +

A +A +A + -

A

=50+60+70+80-5-5-5-5-5-5+1+1+1+1-0

=234

There fore , there are 234 elements in the union of four sets

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Reference

1.Discrete mathematics with application ,Rosen,2007

2.Discrete mathematics with example,Simpison,2002

3.The principle of inclusion and exclusion,brilliant