1. 6  Sets and Set Operations  The Number of Elements in a Finite Set  The Multiplication Principle  Permutations and Combinations Sets and Counting

2. 6.1 Sets and Set Operations

3. Set Terminology and Notation  A set is a well-defined collection of objects. ✦ Sets are usually denoted by upper case letters such as A, B, C, …  The objects of a set are called elements, or members, of a set. ✦ Elements are usually denoted by lower case letters such as a, b, c, …

4. Set Terminology and Notation  The elements of a set may be displayed by listing each element between braces.  For example, using roster notation, the set A consisting of the first three letters of the English alphabet is written A = {a, b, c}  The set B of all letters of the alphabet may be written B = {a, b, c, …, z}

5. Set Terminology and Notation  Another notation commonly used is set-builder notation.  Here, a rule is given that describes the definite property or properties an object x must satisfy to qualify for membership in the set.  For example, the set B of all letters of the alphabet may be written as B = {x | x is a letter of the English alphabet } and is read “B is the set of all elements of x such that x is a letter of the English alphabet.”

6. Set Terminology and Notation  There is also terminology regarding to whether an element belongs to a set or not: ✦ If a is an element of a set A, we write a  A and read “a belongs to A” or “a is an element of A.” ✦ If a is not an element of a set A, we write a  A and read “a does not belong to A” or “a is not an element of A.”

7. Set Equality  Two sets A and B are equal, written A = B, if and only if they have exactly the same elements.

8. Example  Let A, B, and C be the sets  Then, A = B since they both contain exactly the same elements.  Note that the order in which the elements are displayed is immaterial.  Also, A ≠ C since u  A but u  C.  Similarly, we conclude that B ≠ C. Example 1, page 314

9. Subset  If every element of a set A is also an element of a set B, then we say that A is a subset of B and write A  B.  By this definition, two sets A and B are equal if and only if (1) A  B and (2) B  A.

10. Example  Consider again the sets A, B, and C  We find that C  B since every element of C is also an element of B.  Also, A is not a subset of C, written A  C, since u  A but u  C. Example 2, page 315

11. Empty Set  The set that contains no elements is called the empty set and is denoted by Ø.  The empty set, Ø, is a subset of every set.  To see this, observe that Ø has no elements and thus contains no element that is not also in A.

12. Example  List all subsets of the set A = {a, b, c}. Solution  There is one subset containing no elements, Ø.  There are three subsets consisting of one element: {a}, {b}, {c}  There are also three subsets consisting of two elements: {a, b}, {a, c}, {b, c}  Finally there is one subset consisting of three elements, the set A itself.  Therefore, the subsets of A are Ø, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} Example 4, page 315

13. Universal Set  A universal set is the set of all elements of interest in a particular discussion.  It is the largest set in the sense that all sets considered in the discussion of the problem are subsets of the universal set.

14. Example  If the problem at hand is to determine the ratio of female to male students in a college, then the logical choice of a universal set is the set consisting of the whole student body of the college.  If the problem is to determine the ratio of female to male students in the business department of the college, then the set of all students in the business department may be chosen as the universal set. Example 5, page 316

15. Set Union  Let A and B be sets. The union of A and B, written A  B, is the set of all elements that belong to either A or B or both. A  B = {x | x  A or x  B or both }

16. Example  If A = {b, c, d, e} and B = {a, b, c, d}, then A  B = {a, b, c, d, e}. Example 7, page 317 A B e bcd a

17. Set Intersection  Let A and B be sets. The set of elements common to with the sets A and B, written A  B, is called the intersection of A and B. A  B = {x | x  A and x  B}

18. Example  If A = {b, c, d, e} and B = {a, b, c, d}, then A  B = {b, c, d}. Example 8, page 317 A B bcd

19. Complement of a Set  If U is a universal set and A is a subset of U, then the set of all elements in U that are not in A is called the complement of A and is denoted A c. A c = {x | x  U and x  A}

20. Example  If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 4, 6, 8, 10}, then A c = {1, 3, 5, 7, 9}. A U AcAcAcAc

21. Set Complementation  If U is a universal set and A is a subset of U, then a.U c = Øb.Ø c = U c. (A c ) c = A d. A  A c = U e. A  A c = Ø

22. Properties of Set Operations  Let U is a universal set. If A, B, and C are arbitrary subsets of U, then we have the following laws: ✦ Commutative law for union A  B = B  A

23. Properties of Set Operations  Let U is a universal set. If A, B, and C are arbitrary subsets of U, then we have the following laws: ✦ Commutative law for intersection A  B = B  A A  B = B  A

24. Properties of Set Operations  Let U is a universal set. If A, B, and C are arbitrary subsets of U, then we have the following laws: ✦ Associative law for union A  (B  C) = (A  B)  C A  (B  C) = (A  B)  C

25. Properties of Set Operations  Let U is a universal set. If A, B, and C are arbitrary subsets of U, then we have the following laws: ✦ Associative law for intersection A  (B  C) = (A  B)  C A  (B  C) = (A  B)  C

26. Properties of Set Operations  Let U is a universal set. If A, B, and C are arbitrary subsets of U, then we have the following laws: ✦ Distributive law for union A  (B  C) = (A  B)  (A  C) A  (B  C) = (A  B)  (A  C)

27. Properties of Set Operations  Let U is a universal set. If A, B, and C are arbitrary subsets of U, then we have the following laws: ✦ Distributive law for intersection A  (B  C) = (A  B)  (A  C) A  (B  C) = (A  B)  (A  C)

28. Set Operations  Let U is a universal set. If A, B, and C are arbitrary subsets of U, then we have the following laws: ✦ De Morgan’s Laws (A  B) c = A c  B c (A  B) c = A c  B c

29. Set Operations  Let U is a universal set. If A, B, and C are arbitrary subsets of U, then we have the following laws: ✦ De Morgan’s Laws (A  B) c = A c  B c (A  B) c = A c  B c

30. Example  Let U= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} U= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A= {1, 2, 4, 8, 9} B= {3, 4, 5, 6, 8}  Verify by direct computation that (A  B) c = A c  B c. Solution  A  B = {1, 2, 3, 4, 5, 6, 8, 9}, so (A  B) c = {7, 10}.  Moreover, A c = {3, 5, 6, 7, 10} and B c = {1, 2, 7, 9, 10}, so A c  B c = {7, 10}. Example 13, page 319

31. Applied Example: Automobile Options  Let U denote the set of all cars in a dealer’s lot, and let A= {x  U | x is equipped with automatic transmission } A= {x  U | x is equipped with automatic transmission } B= {x  U | x is equipped with air conditioning } C= {x  U | x is equipped with side air bags } C= {x  U | x is equipped with side air bags }  Find an expression in terms of A, B, and C for each of the following sets: ✦ The set of cars with at least one of the given options. ✦ The set of cars with exactly one of the given options. ✦ The set of cars with automatic transmission and side air bags but no air conditioning. Applied Example 14, page 319

32. Applied Example: Automobile Options  Let U denote the set of all cars in a dealer’s lot, and let A= {x  U | x is equipped with automatic transmission } A= {x  U | x is equipped with automatic transmission } B= {x  U | x is equipped with air conditioning } C= {x  U | x is equipped with side air bags } C= {x  U | x is equipped with side air bags }Solution  The set of cars with at least one of the given options is given by A  B  C: A B C U Applied Example 14, page 319

33. Applied Example: Automobile Options  Let U denote the set of all cars in a dealer’s lot, and let A= {x  U | x is equipped with automatic transmission } A= {x  U | x is equipped with automatic transmission } B= {x  U | x is equipped with air conditioning } C= {x  U | x is equipped with side air bags } C= {x  U | x is equipped with side air bags }Solution  The set of cars with exactly one of the given options is given by (A  B c  C c )  (B  C c  A c )  (C  A c  B c ): A C B U Applied Example 14, page 319

34. Applied Example: Automobile Options  Let U denote the set of all cars in a dealer’s lot, and let A= {x  U | x is equipped with automatic transmission } A= {x  U | x is equipped with automatic transmission } B= {x  U | x is equipped with air conditioning } C= {x  U | x is equipped with side air bags } C= {x  U | x is equipped with side air bags }Solution  The set of cars with automatic transmission and side air bags but no air conditioning is given by A  C  B c : A B C U Applied Example 14, page 319

35. 6.2 The Number of Elements in a Finite Set n( Ø )= 0 n(A  B)= n(A) + n(B) – n(A  B) n(A  B  C)= n(A) + n(B) + n(C) – n(A  B) – n(A  C) – n(B  C) + n(A  B  C)

36. Counting Elements of a Set  The number of elements in a finite set is determined by simply counting the elements in the set.  If A is a set, then n(A) denotes the number of elements in A.  For example, if A = {1, 2, 3, …,20} B = {a, b} C = {8} then n(A) = 20, n(B) = 2, and n(C) = 1.  The empty set has no elements in it, so n( Ø ) = 0.

37.  If A and B are disjoint sets, then n(A  B) = n(A) + n(B) Example  If A = {a, c, d} and B = {b, e, f, g}, then we see that A  B = {a, b, c, d, e, f, g} so, n(A  B) = 7.  On the other hand, n(A) + n(B) = 3 + 4 = 7.  Thus, n(A  B) = n(A) + n(B) holds true in this case, because A and B are disjoint sets: A  B = Ø. Counting Elements of a Set Example 1, page 323

38.  In the general case, A and B need not be disjoint, which leads us to the formula n(A  B) = n(A) + n(B) – n(A  B) Example  If A = {a, b, c, d, e} and B = {b, d, f, h}, then, A  B = {a, b, c, d, e, f, h} so n(A  B) = 7.  On the other hand, (A  B) = {b, d}, so n(A  B) = 2.  We also see that n(A) = 5 and n(B) = 4.  Thus, we find that the formula holds true: n(A) + n(B) – n(A  B) = 5 + 4 – 2 = 7 = n(A  B) Counting Elements of a Set Example 2, page 324

39. Applied Example: Consumer Surveys  In a survey of 100 coffee drinkers, it was found that 70 take sugar, 60 take cream, and 50 take both sugar and cream with their coffee. How many coffee drinkers take sugar, cream, or both with their coffee? Solution  Let U be the set of 100 coffee drinkers surveyed, and let A = {x ∈ U | x takes sugar} and B = {x ∈ U | x takes cream}. A = {x ∈ U | x takes sugar} and B = {x ∈ U | x takes cream}.  Then, n(A) = 70, n(B) = 60, and n(A  B) = 50.  The set of coffee drinkers who take sugar, cream, or both with their coffee is given by A  B.  Using the formula for counting elements of a union of not disjoint sets, we find n(A  B) = n(A) + n(B) – n(A  B) = 70 +60 – 50 = 80  Thus, 80 out of the 100 surveyed coffe drinkers take sugar, cream, or both with their coffee. Applied Example 3, page 324

40.  Similar rules can be derived for cases involving more than two sets.  For example, if we have three sets A, B, and C, we find that n(A  B  C) = n(A) + n(B) + n(C) – n(A  B) n(A  B  C) = n(A) + n(B) + n(C) – n(A  B) – n(A  C) – n(B  C) + n(A  B  C) – n(A  C) – n(B  C) + n(A  B  C) Counting Elements of a Set

41. 6.3 The Multiplication Principle

42. The Fundamental Principle of Counting The Multiplication Principle  Suppose there are m ways of performing a task T 1 and n ways of performing a task T 2.  Then, there are mn ways of performing the task T 1 followed by the task T 2.

43.  Three trunk roads connect town A and town B, and two trunk roads connect town B and town C.  Use the multiplication principle to find the number of ways a journey from town A to town C via town B may be completed. Solution  There are three ways of performing the first task: ✦ Going from town A to town B.  Then, there are two ways of performing the second task: ✦ Going from town B to town C.  The multiplication principle says there are 3 · 2 = 6 ways to complete a journey from town A to town C via town B. Example Example 1, page 330

44. Example  Three trunk roads connect town A and town B, and two trunk roads connect town B and town C.  Use the multiplication principle to find the number of ways a journey from town A to town C via town B may be completed. Solution  The six ways to perform the desired task can be seen in the diagram below:  The six possible routes are: (I, a), (I, b), (II, a), (II, b), (III, a), (III, b) A IIIIII C a b B Example 1, page 330

45. Example  Three trunk roads connect town A and town B, and two trunk roads connect town B and town C.  Use the multiplication principle to find the number of ways a journey from town A to town C via town B may be completed. Solution  They can also be seen with the aid of a tree diagram: a (I, a) b (I, b) a (II, a) b (II, b) a (III, a) b (III, b) IIIIII A B C Example 1, page 330

46. Generalized Multiplication Principle  Suppose a task T 1 can be performed in N 1 ways, a task T 2 can be performed in N 2 ways, …, and, finally, a task T n can be performed in N n ways.  Then, the number of ways of performing tasks T 1, T 2, …, T n in succession is given by the product N 1 N 2 ··· N n

47. Example  A coin is tossed 3 times, and the sequence of heads and tails is recorded.  Use the generalized multiplication principle to determine the number of possible outcomes of this activity. Solution  The coin may land in two ways.  Therefore, in three tosses the number of outcomes (sequences) is given by 2 · 2 · 2 = 8 Example 3, page 332

48. Applied Example: Combination Locks  A combination lock is unlocked by dialing a sequence of numbers: ✦ First to the left, then to the right, and to the left again.  If there are ten digits on the dial, determine the number of possible combinations. Solution  There are ten choices for the first number, followed by ten for the second and ten for the third, so by the generalized multiplication principle there are 10 · 10 · 10 = 1000 possible combinations. Applied Example 4, page 332

49. 6.4 Permutations and Combinations 1!= 1 2!= 2 · 1 = 2 3!= 3 · 2 · 1 = 6 4!= 4 · 3 · 2 · 1 = 24 5!= 5 · 4 · 3 · 2 · 1 = 120... 10!= 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 3,628,800

50. Permutations  Given a set of distinct objects, a permutation of the set is an arrangement of these objects in a definite order.  Order matters very often in the real world.  For example, suppose the winning number for the first prize in a raffle is 9237.  Then, the number 2973 cannot be the winner, even though it contains the same digits as 9237.  Here, the set of the four numbers 9, 2, 3, and 7 are arranged in a different order: one arrangement results in a prize winner, the other does not.

51. Example  Find the number of permutations of A = {a, b, c}. Solution  Each permutation of A consists of a sequence of the three letters a, b, c.  Any such sequence can be constructed by filling in each of the three blanks below with one of the three letters:  Now, there are three letters with which we may fill the first blank: a, b, or c.  Having selected a letter for the first blank, there are two letters left for the second blank.  Finally, there is but one letter left to fill the third blank.  Invoking the generalized multiplication principle, we see that there are 3 · 2 · 1 = 6 possible permutations of A. Example 1, page 336

52. Example  Find the number of permutations of A = {a, b, c}. Solution  The tree diagram shows the 6 possible permutations of A: b(a, b, c) c (a, c, b) a (b, a, c) c (b, c, a) a (c, a, b) b (c, b, a) abc Combined outcomes Example 1, page 336

53. n-Factorial  For any natural number n, n!= n(n – 1)(n – 2) · · · · 3 · 2 · 1 0!= 1

54. Example 1!= 1 2!= 2 · 1 = 2 3!= 3 · 2 · 1 = 6 4!= 4 · 3 · 2 · 1 = 24 5!= 5 · 4 · 3 · 2 · 1 = 120... 10!= 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 3,628,800

55. Permutations of n Distinct Objects  The number of permutations of n distinct objects taken r at a time is

56. Example  Compute P(4, 4) and P(4, 2) and interpret the results. Solution  The number of permutations of four objects taken four at a time is 24. Example 3, page 338

57. Example  Compute P(4, 4) and P(4, 2) and interpret the result. Solution  The number of permutations of four objects taken two at a time is 12. Example 3, page 338

58. Applied Example: Committee Selection  Find the number of ways a chairman, a vice-chairman, a secretary, and a treasurer can be chosen from a committee of eight members. Solution  The problem is equivalent to finding the number of permutations of eight distinct objects taken four at a time.  Therefore, there are ways of choosing the four officials from the committee of eight members. Example 5, page 339

59. Permutations of n Objects, Not All Distinct  Given a set of n objects in which n 1 objects are alike and of one kind, n 2 objects are alike and of another kind, …, and n r objects are alike and of yet another kind, so that then the number of permutations of these n objects taken n at a time is given by

60. Example  Find the number of permutations that can be formed from all the letters in the word ATLANTA. Solution  There are seven objects (letters) involved, so n = 7.  Three of them are alike and of one kind (three As), while two of them are alike and of another kind (the two Ts).  Hence, in this case we have n 1 = 3, n 2 = 2, n 3 = 1, and n 4 = 1.  Therefore, there are possible permutations. Example 6, page 340

61. Applied Example: Management Decision  Weaver and Kline, a stock brokerage firm, has received nine inquiries regarding new accounts.  In how many ways can these inquiries be directed to three of the firm’s account executives if each account executive is to handle three inquiries? Solution  If we think of the 9 inquiries as being slots arranged in a row with inquiry 1 on the left and inquiry 9 on the right, then the problem can be thought of as one of filling each slot with a business card from an account executive.  We will use 9 business cards, of which 3 are alike and of one kind, 3 are alike and of another kind, and 3 are alike and of yet another kind. Applied Example 7, page 340

62. Applied Example: Management Decision  Weaver and Kline, a stock brokerage firm, has received nine inquiries regarding new accounts.  In how many ways can these inquiries be directed to three of the firm’s account executives if each account executive is to handle three inquiries? Solution  In this case we have n = 9 and n 1 = n 2 = n 3 = 3.  Therefore, there are possible ways of directing the inquiries. Applied Example 7, page 340

63. Combinations of n Objects  The number of combinations of n distinct objects taken r at a time is given by (where r  n)

64. Example  Compute C(4, 4) and C(4, 2) and interpret the results. Solution  This gives 1 as the number of combinations of four distinct objects taken four at a time. Example 8, page 341

65. Example  Compute C(4, 4) and C(4, 2) and interpret the results. Solution  This gives 6 as the number of combinations of four distinct objects taken two at a time. Example 8, page 341

66. Applied Example: Committee Selection  A Senate investigation subcommittee of four members is to be selected from a Senate committee of ten members.  Determine the number of ways this can be done. Solution  In this case, the order in which the members are selected is unimportant and so the number of ways of choosing the subcommittee is given by C(10, 4).  Hence, there are ways of choosing the subcommittee. Applied Example 9, page 341

67.  Suppose an investor has decided to purchase shares in the stocks of two aerospace companies, two energy development companies, and two electronics companies.  In how many ways can the investor select the group of six companies for the investment from a recommended list of five aerospace companies, three energy development companies, and four electronics companies? Applied Example: Investment Options Applied Example 12, page 343

68. Applied Example: Investment Options Solution  There are C(5, 2) ways to select the aerospace companies, C(3, 2) ways to select the energy development companies, and C(4, 2) ways to select the electronics companies as investments.  By the generalized multiplication principle, there are ways of selecting the group of six companies. Applied Example 12, page 343

69. End of Chapter