Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 1 of 93 Chapter 5 Sets and Counting.

Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 1 of 93

Chapter 5

Sets and Counting


Outline

5.1 Sets5.2 A Fundamental Principle of Counting5.3 Venn Diagrams and Counting5.4 The Multiplication Principle5.5 Permutations and Combinations5.6 Further Counting Problems5.7 The Binomial Theorem5.8 Multinomial Coefficients and Partitions


5.1 Sets

1. Sets and Elements

2. Union and Intersection

3. Subset and Empty Set

4. Universal Set and Complement


Sets and Elements

A set is any collection of objects. The objects, which may be countries, cities, years, numbers, letters, or anything else, are called the elements of the set. A set is often specified by a listing of its elements inside a pair of braces. A set may also be specified by giving a description of its elements.


1. The set of the first six letters of the alphabet is {a, b, c, d, e, f}.

2. {2, 4, 6, 8, 10} is the set {the even numbers between 1 and 11}.

3. The graph {(a,b) where b = a2} is a set with infinitely many elements.

Example Sets and Elements


Example Sets and Elements (2)

Let A = {years from 1991 to 2004 in which unemployment is at least 6%}.

Let B = {years from 1991 to 2004 in which inflation is at least 3%}.


Example Sets and Elements (3)

Using the table, the two sets are

A = {1991, 1992, 1993, 1994, 2003} and

B = {1991, 1992, 1993, 1996, 2000}.


Union and Intersection

The union of A and B, denoted AB, is the set consisting of all those elements that belong to either A or B or both. The intersection of A and B, denoted AB, is the set consisting of those elements that belong to both A and B.


Let A = {1991, 1992, 1993, 1994, 2003} and

B = {1991, 1992, 1993, 1996, 2000} from the previous example. Find

AB

AB

Example Union and Intersection

= {1991,1992,1993,1994,1996,2000,2003}

= {1991,1992,1993}


Subset and Empty Set

A set B is called a subset of A provided that every element of B is also an element of A.

The set that contains no elements at all is the empty set (or null set) and is written as . The empty set is a subset of every set.


List all possible subsets of {a, b, c}.

{a}, {b}, and {c}

{a, b}, {a, c}, and {b, c}

{a, b, c}

Example Subset and Empty Set


Universal Set and Complement

The set U that contains all the elements of the sets being discussed is called a universal set (for the particular problem).

If A is a subset of U, the set of elements in U that are not in A is called the complement of A, denoted by A'.


Example Universal Set and Complement

Let U = {a,b,c,d,e,f,g}, S = {a,b,c} and T = {a,c,d}.Find

S

T

S T

= {d,e,f,g}

= {b,e,f,g}

= {b,d,e,f,g}


A set is a collection of objects. Each object is called an element of the set. The empty set is the set containing no objects. The union of two sets is the set consisting of all elements that belong to at least one of the sets. The intersection of two sets is the set consisting of all elements that belong to both of the sets.

Summary Section 5.1 - Part 1


Set A is a subset of set B if every element of set A is also an element of set B. In each situation or problem, all sets are considered to be subsets of a universal set. The set of all elements in the universal set that do not belong to the set A is called the complement of A, denoted A'.



5.2 A Fundamental Principle of Counting

1. Number of Elements in A

2. Inclusion-Exclusion Principle

3. Venn Diagram

4. De Morgan's Laws


Number of Elements in A

If S is any set, we will denote the number of elements in S by n(S).

For example,

if S = {1,7,11}, then n(S) = 3;

if S = , then n(S) = 0.


Inclusion-Exclusion Principle

Inclusion-Exclusion Principle Let S and T be sets. Then

n(ST) = n(S) + n(T) - n(ST).


Example Inclusion-Exclusion Principle

In the year 2000, Executive magazine surveyed the presidents of the 500 largest corporations in the US. Of these 500 people, 310 had degrees (of any sort) in business, 238 had undergraduate degrees in business, and 184 had graduate degrees in business. How many presidents had both undergraduate and graduate degrees in business?


Example Inclusion-Exclusion Principle (2)

Let S = {presidents with an undergraduate degree in business} and T = {presidents with a graduate degree in business}.

ST = {presidents with degrees (of any sort) in business}

ST = {presidents with both undergraduate and graduate degrees in business}


Example Inclusion-Exclusion Principle (3)

n(S) = 238

n(T) = 184

n(ST) = 310

n(ST) = n(S) + n(T) - n(ST)

310 = 238 + 184 - n(ST)

n(ST) = 238 + 184 - 310 = 112


Venn Diagram

A Venn diagram is a drawing that represents sets geometrically. To construct a Venn diagram, draw a rectangle and view its points as the elements of U. Then draw a circle inside the rectangle for each set. The circles should overlap. View the points inside the circles as elements of each set.


Example Venn Diagram

Draw a Venn diagram with three sets, R, S and T. Shade the area that represents RST, RST and SR'T '.


Example Venn Diagram (2)


De Morgan's Laws

De Morgan's Laws Let S and T be sets. Then

(ST)' = S'T',

(ST)' = S'T'.


Example De Morgan's Laws

Verify (ST)' = S 'T ' using Venn diagrams.


Example De Morgan's Laws (2)


Example De Morgan's Laws (3)


The inclusion-exclusion principle says that the number of elements in the union of two sets is the sum of the number of elements in each set minus the number of elements in their intersection.



A Venn diagram consists of a rectangle containing overlapping circles and is used to depict relationships among sets. The rectangle represents the universal set and the circles represent subsets of the universal set. De Morgan's laws state that the complement of the union (intersection) of two sets is the intersection (union) of their complements.



5.3 Venn Diagrams and Counting

1. Basic Regions

2. Basic Regions and Counting


Basic Regions

Each Venn diagram divides the universal set into non-overlapping regions called basic regions.


Example Basic Regions

Show the basic regions for the Venn diagram with two sets and with three sets.


Example Basic Regions and Counting

For the survey of the presidents of the 500 largest corporations in the US where 310 had degrees (of any sort) in business, 238 had undergraduate degrees in business, and 184 had graduate degrees in business: a. Draw a Venn diagram and determine the number of elements in each basic region;b. Determine the number of presidents having exactly one degree in business.



n(I) = n(U) - n(ST)

= 500 - 310 = 190

n(III) = n(ST) = 112

n(II) = n(S) - n(ST)

= 238 - 112 = 126n(IV) = n(T) - n(ST) = 184 - 112 = 72



The number of presidents with exactly one business degree corresponds to the shaded region. Adding the number of people in the two shaded basic regions gives 126 + 72 = 198.


Summary Section 5.3

The universal set can be divided up into a number of non-overlapping regions called basic regions. These regions can be used to determine the number of elements in a variety of subsets of the universal set.


5.4 The Multiplication Principle

1. Tree Diagram

2. Multiplication Principle

3. Generalized Multiplication Principle


Tree Diagram

A tree diagram is a graph showing possibilities for an event having choices along the way. Suppose a rat in a maze starts at point A. There are five possible routes to get from point A to point B and 3 possible routes to get from point B to the final destination, point C. Represent the possible choices using a tree diagram.


Tree Diagram (2)

Possible paths:

1a, 1b, 1c

2a, 2b, 2c

3a, 3b, 3c

4a, 4b, 4c

5a, 5b, 5c

Total possible paths = 15


Multiplication Principle

Multiplication Principle Suppose that a task is composed of two consecutive operations. If operation 1 can be performed in m ways and, for each of these, operation 2 can be performed in n ways, then the complete task can be performed in mn ways.


Example Multiplication Principle

Use the multiplication principle to determine the number of different paths a rat can take going from point A to point C if there are five possible routes to get from point A to point B and 3 possible routes to get from point B to point C.

There are 5 possible routes for task 1 and 3 possible routes for task 2 so there are 53 = 15 possible routes. This agrees with the tree diagram drawn previously.


Generalized Multiplication Principle

Generalized Multiplication Principle Suppose that a task is composed of t operations performed consecutively. Suppose operation 1 can be performed in m1 ways; for each of these, operation 2 can be performed in m2 ways; for each of these, operation 3 can be performed in m3 ways; and so forth. Then the complete task can be performed in

m1m2m3mt ways.


Example Generalized Principle

A corporation has a board of directors consisting of 10 members. The board must select from among its members a chairperson, vice chairperson, and secretary. In how many ways can this be done?


Example Generalized Principle (2)

The operations are:

1. Select chairperson - there are 10 ways (people) to do this.

2. Select a vice chairperson - there are 9 ways (9 people left after chairperson is chosen) to do this.

3. Select a secretary - there are 8 ways to do this.

Total number of ways to complete task is 1098 = 720 ways.


Summary Section 5.4

The multiplication principle states that the number of ways a sequence of several independent operations can be performed is the product of the number of ways each individual operation can be performed.


5.5 Permutations and Combinations

1. Permutation2. Combination

3. Formula for P(n,r)4. Factorial

5. Formula for C(n,r)


Permutation

Given a set of n objects, a permutation of n objects taken r at a time is an arrangement of r of the n objects in a specific order.

The notation for a permutation is

P(n,r) = the number of permutations of n objects taken r at a time.


Example Permutation

How many words (strings of letters) of two distinct letters can be formed from the letters {a, b, c}?

The number of objects is n = 3.

The number of objects taken at a time is r = 2.

The possibilities are ab, ac, ba, bc, ca and cb.

P(3,2) = 6.

Note: the word ab is different from the word ba.


Combination

Given a set of n objects, a combination of n objects taken r at a time is a selection of r of the n objects with order disregarded.

The notation for a combination is

C(n,r) = the number of combinations of n objects taken r at a time.


Example Combination

How many two-member teams can be formed from a group that has three members a, b, and c?



The possibilities are ab, ac and bc.

C(3,2) = 3.

Note: the team ab is the same as the team ba.


Formula for P(n,r)

P(n,1) = n

P(n,2) = n(n - 1)

P(n,3) = n(n - 1)(n - 2)

P(n,r) = n(n - 1)(n - 2) (n - r + 1)

r factors


Example Formula for P(n,r)

Eight horses are entered in a race in which a first, second, and third prize will be awarded. Assuming no ties, how many different outcomes are possible?



The number of outcomes is P(8,3) = 876 = 336.


Factorial

1! = 12! = 21 = 23! = 321 = 64! = 4321 = 24n! = n(n - 1)(n - 2) (1)Note: n! is read as “n factorial.”


Formula for C(n,r)

, 1 1,

! 1 1

P n r n n n rC n r

r r r


Example Formula for C(n,r)

The board of directors of a corporation has 10 members. In how many ways can they choose a committee of 3 board members to negotiate a merger?



Number of possible committees is

10 9 810,3 120.

3 2 1C


Summary Section 5.5

The number of ordered arrangements, each called a permutation, of n objects taken r at a time is

P(n,r) = n(n - 1)(n - 2) (n - r + 1). The number of unordered arrangements, each called a combination, of n objects taken r at a time is

, 1 1, .

! 1 1

P n r n n n rC n r

r r r


5.6 Further Counting Problems

1. Tossing Coins2. Routes3. Balls Drawn From an Urn


Example Tossing Coins

An experiment consists of tossing a coin 10 times and observing the sequence of heads and tails.

a. How many different outcomes are possible?

b. How many different outcomes have exactly two heads?

c. How many different outcomes have at most two heads?

d. How many different outcomes have at least two heads?


Example Tossing Coins (a)

A possible outcome is

H T H T T T H T H T

where H is heads and T is tails.

Each coin has two possible outcomes.

By the generalized multiplication principle, the total number of possible outcomes is

2222222222 = 210 = 1024.


Example Tossing Coins (b)

A possible outcome with 2 heads is

H T H T T T T T T T.

The 2 heads must be placed in 2 of the 10 possible positions.

The number of outcomes with 2 heads is

10 910,2 45.

2 1C


Example Tossing Coins (c)

At most 2 heads means there can be 0 heads or 1 head or 2 heads.

There is only 1 possible outcome with no heads and that is if all 10 coins are tails.

There are C(10,1) = 10 possible outcomes with 1 head.

There are C(10,2) = 45 possible outcomes with 2 heads.

Therefore, there are 1 + 10 + 45 = 56 possible outcomes with at most two heads.


Example Tossing Coins (d)

At least 2 heads means there can not be 0 heads or 1 head.

There are 1 + 10 = 11 possible outcomes with 0 or 1 head.

There are 1024 possible outcomes total.

So, there are 1024 - 11 = 1013 possible outcomes with at least 2 heads.


Example Routes

A tourist in a city wants to walk from point A to point B shown in the maps below. What is the total number of routes (with no backtracking) from A to B?


Example Routes (2)

If S is walking a block south and E is walking a block east, the two possible routes shown in the maps could be designated as SSEEESE and ESESEES.


Example Routes (3)

All routes can be designated as a string of 7 letters, 3 of which will be S and 4 E.

Selecting a route is the same as selecting where in the string the 3 S’s will be placed.

Therefore the total number of possible routes is

7 6 57,3 35.

3 2 1C


Example Balls Drawn From an Urn

An urn contains 25 numbered balls, of which 15 are red and 10 are white. A sample of 5 balls is to be selected.

a. How many different samples are possible?

b. How many different samples contain all red balls?

c. How many samples contain 3 red balls and 2 white balls?


Example Balls Drawn From an Urn (a)

A sample is just an unordered selection of 5 balls out of 25.

25 24 23 22 2125,5 53,130

5 4 3 2 1C


Example Balls Drawn From an Urn (b)

To form a sample of all red balls we must select 5 balls from the 15 red ones.

15 14 13 12 1115,5 3,003

5 4 3 2 1C


Example Balls Drawn From an Urn (c)

To form the sample with 3 red balls and 2 white balls, we must

Operation 1: select 3 red balls from 15 red balls,

Operation 2: select 2 white balls from 10 white balls.

Using the multiplication principle, this gives

15 14 13 10 915,3 10,2 20,475.

3 2 1 2 1C C


5.7 The Binomial Theorem

1. Formulas for C(n,r)2. Binominal Coefficient3. Binomial Theorem4. Number of Subsets


Formulas for C(n,r)

, 1 1,

! 1 1

!,

! !

, ,

P n r n n n rC n r

r r r

nC n r

r n r

C n r C n n r


Example Routes

Work the route problem covered previously by selecting where in the string of length 7 the 4 E’s will be placed instead of the 3 S’s.

Therefore the total number of possible routes is

7 6 5 47,4 35.

4 3 2 1C

Notice that C(7,4) = C (7,3).


Binominal Coefficient

Another notation for C(n,r) is . n

r

n

r

is called a binominal coefficient.


Example Binominal Coefficient


Binominal Theorem

Binomial Theorem

1 2 2

1

0 1 2

1

n n n n

n n

n n n

n n

n n

x y x x y x y

xy y


Example Binominal Theorem

Expand (x + y )5.

(x + y )5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5

5 5 5 5 5 5

0 1 2 3 4 51 5 10 10 5 1


Number of Subsets

A set with n elements has 2n subsets.


Example Number of Subsets

A pizza parlor offers a plain cheese pizza to which any number of six possible toppings can be added. How many different pizzas can be ordered?

Ordering a pizza requires selecting a subset of the 6 possible toppings.

There are 26 = 64 different pizzas.



C(n,r) is also denoted by . The formula C(n,r) = C(n,n - r) simplifies the computation of C(n,r) when r is greater than n/2. The binomial theorem states that

n

r

1 2 2

1

0 1 2

1.

n n n n

n n

n n n

n n

n n

x y x x y x y

xy y



A set with n elements has 2n subsets.


5.8 Multinomial Coefficients and Partitions

1. Ordered Partition

2. Number of Ordered Partitions

3. Unordered Partition

4. Number of Unordered Partitions


Ordered Partition

Let S be a set of n elements. An ordered partition of S of type (n1,n2,…,nm) is a decomposition of S into m subsets (given in a specific order) S1, S2,…,Sm, where no two of these intersect and where n(S1) = n1, n(S2) = n2, … , n(Sm) = nm and n1 + n2 +… + nm = n.


List all ordered partitions of S = {a, b, c, d} of type (1,1,2).

({a},{b},{c,d}) ({a},{c},{b,d}) ({a},{d},{b,c}) ({b},{a},{c,d}) ({b},{c},{a,d}) ({b},{d},{a,c}) ({c},{a},{b,d}) ({c},{b},{a,d}) ({c},{d},{a,b}) ({d},{a},{b,c}) ({d},{b},{a,c}) ({d},{c},{a,b})

Example Ordered Partition


Number of Ordered Partitions

The number of ordered partition of type (n1,n2,…,nm) for a set with n elements is

1 2 1 2

!.

, , , ! ! !m m

n nn n n n n n

The above is called the multinomial coefficient.


A work crew consists of 12 construction workers, all having the same skills. A construction job requires 4 welders, 3 concrete workers, 3 heavy equipment operators, and 2 bricklayers. In how many ways can the 12 workers be assigned to the required tasks?

Example Ordered Partition


Each assignment of jobs corresponds to an ordered partition of the type (4,3,3,2).

The number of ordered partitions is

Example Ordered Partition (2)

12 12!277,200.

4,3,3,2 4!3!3!2!


Unordered Partition

Let S be a set of n elements. An unordered partition of S of type (n1,n2,…,nm) is a decomposition of S into m subsets S1, S2,…,Sm, where no two of these intersect and where n(S1) = n1, n(S2) = n2, … , n(Sm) = nm and n1 + n2 +… + nm = n.


List all unordered partitions of S = {a, b, c, d} of type (1,1,2).

({a},{b},{c,d}) ({a},{c},{b,d}) ({a},{d},{b,c}) ({b},{c},{a,d}) ({b},{d},{a,c}) ({c},{d},{a,b})

Example Unordered Partition


Number of Unordered Partitions

The number of unordered partitions of type (n1,n1,…,n1) for a set with n elements is

1

1 !

! !m

n

m n


A work crew consists of 12 construction workers, all having the same skills. In how many ways can the workers be divided into 4 groups of 3?

Example Unordered Partition

4

1 12!15,400.

4! 3!



Let S be a set of n elements, and suppose that n = n1 + n2 +… + nm where each number in the sum is a positive integer. Then the number of ordered partitions of S into subsets of sizes n1, n2, …, nm

is

1 2 1 2

!.

, , , ! ! !m m

n nn n n n n n



Let S be a set of n elements, where n = m r. Then the number of unordered partitions of S into m subsets of size r is

1 !

.! !

m

n

m r

Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 1 of 93 Chapter 5 Sets and Counting.

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Transcript of Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 1 of 93 Chapter 5 Sets and Counting.