College Algebra Sixth Edition James Stewart  Lothar Redlin  Saleem Watson

1. College Algebra Sixth Edition James StewartLothar RedlinSaleem Watson

2. Probability and Statistics 9

3. Counting 9.1

4. The Fundamental Counting Principle

5. Overview • Suppose there are three towns: • Ashbury, Brampton, and Carmichael • They are located in such a way that • Two roads connect Ashbury to Brampton • Three roads connect Brampton to Carmichael

6. Overview • How many different routes can one take to travel from Ashbury to Carmichael via Brampton?

7. Overview • The key to answering this question is to consider the problem in stages. • At the first stage (from Ashbury to Brampton) there are two choices. • For each of these choices, there are three choices at the second stage (from Brampton to Carmichael).

8. Overview • Thus, the number of different routes is 2 × 3 = 6. • These routes areconvenientlyenumerated bya tree diagramas in the figure.

9. The Fundamental Counting Principle • The method that we used to solve this problem leads to the following principle. • The Fundamental Counting Principle: • Suppose that two events occur in order. • If the first can occur in m ways and second in n ways(after the first has occurred). • Then the two events can occur in order in m× n ways.

10. Overview • There is an immediate consequence of this principle for any number of events: • If E1, E2, …, Ek are events thatoccur in order • And if E1 can occur in n1 ways,E2 in n2 ways, and so on • Then the events can occur in order inn1× n2 × … × nk ways

11. E.g. 1—Using the Fundamental Counting Principle • An ice-cream store offers three types of cones and 31 flavors. • How many different single-scoop ice-cream cones is it possible to buy at this store?

12. E.g. 1—Using the Fundamental Counting Principle • There are two choices: • Type of cone • Flavor of ice cream • At the first stage, we choose a type of cone. • And at the second stage, we choose a flavor.

13. E.g. 1—Using the Fundamental Counting Principle • We can think of the different stages as boxes:

14. E.g. 1—Using the Fundamental Counting Principle • The first box can be filled in three ways, and the second can be filled in 31 ways: • Thus, by the Fundamental Counting Principle there are 3 × 31 = 93 ways of choosing a single-scoop ice-cream cone at this store.

15. E.g. 2—Using the Fundamental Counting Principle • In a certain state, automobile license plates display three letters followed by three digits. • How many such plates are possible if repetition of the letters • Is allowed? • Is not allowed?

16. Example (a) E.g. 2—Using the Fundamental Counting Principle • There are six selection stages, one for each letter or digit on the license plate. • As in the preceding example, we sketcha box for each stage:

17. Example (a) E.g. 2—Using the Fundamental Counting Principle • At the first stage, we choose a letter(from 26 possible choices) • At the second stage, we choose another letter(again from 26 choices) • At the third stage, we choose another letter(26 choices)

18. At the fourth stage, we choose a digit(from 10 possible choices) At the fifth stage, we choose a digit(again from 10 choices) At the sixth stage, we choose another digit(10 choices) Example (a) E.g. 2—Using the Fundamental Counting Principle

19. Example (a) E.g. 2—Using the Fundamental Counting Principle • By the Fundamental Counting Principle,the number of possible license plates is • 26 × 26 × 26 × 10 × 10 × 10 = 17,576,000

20. Example (b) E.g. 2—Using the Fundamental Counting Principle • If repetition of letters is not allowed, then we can arrange the choices as follows:

21. Example (b) E.g. 2—Using the Fundamental Counting Principle • At the first stage, we have 26 letters to choose from. • But once the first letter has been chosen, there are only 25 letters to choose fromat the second stage. • Once the first two letters have beenchosen, 24 letters are left to choose fromfor the third stage. • The digits are chosen as before.

22. Example (b) E.g. 2—Using the Fundamental Counting Principle • By the Fundamental Counting Principle,the number of possible license plates is • 26 × 25 × 24 × 10 × 10 × 10 = 15,600,000

23. The Number of Subsets of a Set • Let S be a set with n elements. A subset of S can be chosen by making one of two choices for each element: • We can choose the element to be in or out of A. • By the Fundamental Counting Principle the total number of different subsets is 2  2  . . .  2, where there are n factors.

24. The Number of Subsets of a Set • A set with n elements has 2n different subsets.

25. E.g. 3—Finding the Number of Subsets of a Set • A pizza parlor offers the basic cheese pizza and a choice of 16 toppings. • How many different kinds of pizzas can beordered at this pizza parlor?

26. E.g. 3—Finding the Number of Subsets of a Set • We need the number of possible subsets of the 16 toppings. • Including the empty set, which correspondsto a plain cheese pizza. • Thus, 216 = 65,536 different pizzas can beordered.

27. Counting Permutations

28. Permutations • A permutation of a set of distinct objects is an ordering of these objects. • For example, some permutations of the letters ABCD are ABDC BACD DCBA DABC • How many such permutations are possible?

29. Permutations • Since there are: • Four choices for the first position, • Three for the second(after the first has been chosen), • Two for the third(after the first two have been chosen), • Only one choice for the fourth letter • By the Fundamental Counting Principle the number of possible permutations is 4 × 3 × 2 × 1 = 4! = 24

30. Permutations • This same reasoning with 4 replaced by n leads to the following observation. • The number of permutations of n objects is n!.

31. Permutations • How many permutations consisting of two letters can be made from these same four letters? • Some of these permutations areAB AC BD DB

32. Permutations • Again, there are four choices for the first position, three for the second, two for the third, and only one choice for the fourth. • By the Fundamental Counting Principle, the number of such permutations is 4 × 3 = 12

33. Permutations • In general, if a set has n elements, then the number of ways of ordering r elements from the set is denoted P(n, r). • This is called the number of permutations of n objects taken r at a time.

34. Permutations of n Objects Taken r at a Time • The number of permutations of n objects taken r at a time is

35. Permutations of n Objects Taken r at a Time There are n objects and r positions to placethem in. Thus, there are n choices for the first position,n – 1 choices for the second position, and so on. The last position can be filled in n – r + 1 ways.

36. Permutations of n Objects Taken r at a Time • By the Fundamental Counting Principle,P(n, r) = n(n – 1)(n – 2)…(n – r + 1) • This formula can be written more compactlyusing factorial notation:

37. E.g. 4—Finding the Number of Permutations • There are six runners in a race that is completed with no tie. • In how many different ways can the race be completed? • In how many different ways can first, second, and third place be decided?

38. Example (a) E.g. 4—Finding the Number of Permutations • The number of ways to complete the race is the number of permutations of the six • runners: 6! = 720.

39. Example (b) E.g. 4—Finding the Number of Permutations • The number of ways in which the first three positions can be decided is

40. E.g. 5—Finding the Number of Permutations • A club has nine members. • In how many ways can a president,vice president, and secretary bechosen from the members of this club? • We need the number of ways of selectingthree members, in order, for these positionsfrom the nine club members.

41. E.g. 5—Finding the Number of Permutations • This number is

42. E.g. 6—Finding the Number of Permutations • From 20 raffle tickets in a hat, four tickets are to be selected in order. • The holder of the first ticket wins a car, • The second a motorcycle, • The third a bicycle, • And, the fourth a skateboard.

43. E.g. 6—Finding the Number of Permutations • In how many different ways can these prizes be awarded? • The order in which the tickets are chosen determines who wins each prize. • So, we need to find the number of ways of selecting four objects, in order, from 20 objects(the tickets).

44. E.g. 6—Finding the Number of Permutations • This number is

45. Distinguishable Permutations

46. Distinguishable Permutations • If we have a collection of ten balls, each a different color, then the number of permutations of these balls is P(10, 10) = 10!. • If all ten balls are red, then we have justone distinguishable permutation becauseall the ways of ordering these balls lookexactly the same.

47. Distinguishable Permutations • In general, in considering a set of objects, some of which are the same kind: • Then, two permutations are distinguishableif one cannot be obtained from the other byinterchanging the positions of elements of the same kind.

48. Distinguishable Permutations • For example, if we have ten balls, of which • Six are red • The other four are each a different color • Then, how many distinguishable permutations are possible?

49. Distinguishable Permutations • The key point here is that balls of the same color are not distinguishable. • So each arrangement of the red balls, keeping all the other balls fixed, gives essentially the same permutation. • There are 6! rearrangements of the red ball for each fixed position of the other balls.

50. Distinguishable Permutations The key point here is that balls of the same color are not distinguishable. Thus, the total number of distinguishable permutations is 10!/6!. The same type of reasoning gives the followinggeneral rule.