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4 th EDITION

College Algebra & Trigonometry and Precalculus. 4 th EDITION. 11.6. Counting Theory. Fundamental Principle of Counting Permutations Combinations Distinguishing Between Permutations and Combinations. Fundamental Principle of Counting.

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4 th EDITION

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  1. College Algebra & Trigonometry and Precalculus 4th EDITION

  2. 11.6 Counting Theory Fundamental Principle of Counting Permutations Combinations Distinguishing Between Permutations and Combinations

  3. Fundamental Principle of Counting If there are 3 roads from Albany to Baker and 2 roads from Baker to Creswich, in how many ways can one travel from Albany to Creswich by way of Baker?

  4. Fundamental Principles of Counting For each of the 3 roads from Albany to Baker, there are 2 different roads from Baker to Creswich. Hence, there are 6 different ways to make the trip, as shown in the tree diagram.

  5. Fundamental Principles of Counting In this situation, each choice of road is an example of an event. Two eventsare independent events if neither influences the outcome of the other. The opening example illustrates the fundamental principle of counting with independent events.

  6. Fundamental Principle of Counting If n independent events occur, with m1 ways for event 1 to occur, m2 ways for event 2 to occur, and mn ways for event n to occur, then there are m1·m2· · · · · mn different ways for all n events to occur.

  7. USING THE FUNDAMENTAL PRINCIPLE OF COUNTING Example 1 A restaurant offers a choice of 3 salads, 5 main dishes, and 2 desserts. Use the fundamental principle of counting to find the number of different 3-course meals that can be selected. Solution Three events are involved: selecting a salad, selecting a main dish, and selecting a dessert. The first event can occur in 3 ways, the second event can occur in 5 ways, and the third event can occur in 2 ways. Thus, there are possible meals.

  8. USING THE FUNDAMENTAL PRINCIPLE OF COUNTING Example 2 A teacher has 5 different books that he wishes to arrange in a row. How many different arrangements are possible? Solution Five events are involved: selecting a book for the first spot, selecting a book for the second spot, and so on. For the first spot the teacher has 5 choices. After a choice has been made, the teacher has 4 choices for the second spot.

  9. USING THE FUNDAMENTAL PRINCIPLE OF COUNTING Example 2 Continuing in this manner, there are 3 choices for the third spot, 2 for the fourth spot, and 1 for the fifth spot. By the fundamental principle of counting, there are arrangements.

  10. In using the fundamental principle of counting, products such as occur often. We use the symbol n! (read “n-factorial”), for any counting number n, as follows. Thus, is written 5! and is written 3!. By the definition of n!, n[(n – 1)!] =n! for all natural numbers n ≥ 2. It is convenient to have this relation hold also for n = 1, so, by definition,

  11. ARRANGING r OF n ITEMS (r < n) Example 3 Suppose the teacher in Example 2 wishes to place only 3 of the 5 books in a row. How many arrangements of 3 books are possible? Solution The teacher still has 5 ways to fill the first spot, 4 ways to fill the second spot, and 3 ways to fill the third. Since only 3 books will be used, there are only 3 spots to be filled (3 events) instead of 5, with arrangements.

  12. Permutations Since each ordering of three books is considered a different arrangement, the number 60 in the preceding example is called the number of permutations of 5 things taken 3 at a time, written P(5, 3) = 60. The number of ways of arranging 5 elements from a set of 5 elements, written P(5, 5) = 120, was found in Example 2.

  13. Permutations A permutation of n elements taken r at a time is one of the arrangements of r elements from a set of n elements. Generalizing from the examples provided, the number of permutations of n elements taken r at a time, denoted by P(n, r), is

  14. Permutations of n Elements Taken r at a Time If P(n, r) denotes the number of permutations of n elements taken r at a time, with r ≤ n, then Alternative notations for P(n, r)are and

  15. USING THE PERMUTATIONS FORMULA Example 4 Find each value. (a) The number of permutations of the letters L, M, and N Solution (a) By the formula for P(n, r), with n = 3 and r = 3,

  16. USING THE PERMUTATIONS FORMULA Example 4 Find each value. (a) The number of permutations of the letters L, M, and N Solution As shown in the tree diagram here, the 6 permutations are LMN, LNM, MLN, MNL, NLM, NML.

  17. USING THE PERMUTATIONS FORMULA Example 4 Find each value. (b) The number of permutations of 2 of the letters L, M, and N Solution (b) Find P(3, 2). This result is the same as the answer in part (a). After the first two choices are made, the third is already determined since only one letter is left.

  18. USING THE PERMUTATIONS FORMULA Example 5 Suppose 8 people enter an event in a swim meet. In how many ways could the gold, silver, and bronze medals be awarded? Solution Using the fundamental principle of counting, there are 3 events, giving choices. We can also use the formula for P(n, r) to get the same result. There are

  19. USING THE PERMUTATIONS FORMULA Example 6 In how many ways can 6 students be seated in a row of 6 desks? Solution Use P(n, r) with n = 6 and r = 6 to get

  20. Combinations In Example 3 we saw that there are 60 ways that a teacher can arrange 3 of 5 different books in a row. That is, there are 60 permutations of 5 things taken 3 at a time. Suppose now that the teacher does not wish to arrange the books in a row but rather wishes to choose, without regard to order, any 3 of the 5 books to donate to a book sale.

  21. Combinations In how many ways can the teacher do this? The number 60 counts all possible arrangements of 3 books chosen from 5. The following 6 arrangements, however, would all lead to the same set of 3 books being given to the book sale. mystery-biography-textbook biography-textbook-mystery mystery-textbook-biography textbook-biography-mystery biography-mystery-textbook textbook-mystery-biography

  22. Combinations The list shows 6 different arrangements of 3 books but only one set of 3 books. A subset of items selected without regard to order is called a combination. The number of combinations of 5 things taken 3 at a time is written

  23. Note This combinations notation also represents the binomial coefficient defined in Section 7.4. That is, binomial coefficients are the combinations of n elements chosen r at a time.

  24. Combinations To evaluate start with the permutations of5 things taken 3 at a time. Since order does not matter, and each subset of 3 items from the set of 5 items can have its elements rearranged in ways, we find by dividing the number of permutations by 3!, or

  25. Combinations The teacher can choose 3 books for the book sale in 10 ways. Generalizing this discussion gives the following formula for the number of combinations of n elements taken r at a time:

  26. Combinations An alternative version of this formula is found as follows. This version is most useful for calculation and is the one we used earlier to calculate binomial coefficients.

  27. Combinations of n Elements Taken r at a time If represents the number of combinations of n elements taken r at a time, with r ≤ n, then

  28. Note The formula for C(n, r) given in the previous slide is equivalent to the binomial coefficient formula given in Section 7.4, with denominator factors rearranged.

  29. USING THE COMBINATIONS FORMULA Example 7 How many different committees of 3 people can be chosen from a group of 8 people? Solution Since a committee is an unordered set, use combinations. There are

  30. USING THE COMBINATIONS FORMULA Example 8 Three stockbrokers are to be selected from a group of 30 to work on a special project. (a) In how many different ways can the stockbrokers be selected? Solution (a) Here we wish to know the number of 3-element combinations that can be formed from a set of 30 elements. (We want combinations, not permutations, since order within the group does not matter.) There are 4060 ways to select the project group.

  31. USING THE COMBINATIONS FORMULA Example 8 Three stockbrokers are to be selected from a group of 30 to work on a special project. (b) In how many ways can the group of 3 be selected if a particular stockbroker must work on the project? Solution (b) Since 1 broker has already been selected for the project, the problem is reduced to selecting 2 more from the remaining 29 brokers. In this case, the project group can be selected in 406 ways.

  32. Distinguishing Between Permutations and Combinations Students often have difficulty determining whether to use permutations or combinations in solving problems. The following chart lists some of the similarities and differences between these two concepts.

  33. Distinguishing Between Permutations and Combinations

  34. DISTINGUISHING BETWEEN PERMUTATIONS AND COMBINATIONS Example 9 Should permutations or combinations be used to solve each problem? (a) How many 4-digit codes are possible if no digits are repeated? Solution (a) Since changing the order of the 4 digits results in a different code, permutations should be used.

  35. DISTINGUISHING BETWEEN PERMUTATIONS AND COMBINATIONS Example 9 Should permutations or combinations be used to solve each problem? (b) A sample of 3 light bulbs is randomly selected from a batch of 15 bulbs. How many different samples are possible? Solution (b) The order in which the 3 light bulbs are selected is not important. The sample is unchanged if the items are rearranged, so combinations should be used.

  36. DISTINGUISHING BETWEEN PERMUTATIONS AND COMBINATIONS Example 9 Should permutations or combinations be used to solve each problem? (c) In a basketball tournament with 8 teams, how many games must be played so that each team plays every other team exactly once? Solution (c) Selection of 2 teams for a game is an unordered subset of 2 from the set of8 teams. Use combinations.

  37. DISTINGUISHING BETWEEN PERMUTATIONS AND COMBINATIONS Example 9 Should permutations or combinations be used to solve each problem? (d) In how many ways can 4 stockbrokers be assigned to 6 offices so that each broker has a private office? Solution (d) The office assignments are an ordered selection of 4 offices from the 6 offices. Exchanging the offices of any 2 brokers within a selection of 4 offices gives a different assignment, so permutations should be used.

  38. Distinctions Between Permutations and Combinations To illustrate the distinctions between permutations and combinations in another way, suppose we want to select 2 cans of soup from 4 cans on a shelf: noodle (N), bean (B), mushroom (M), and tomato (T). As shown in the first diagram on the next slide, there are 12 ways to select 2 cans from the 4 cans if order matters (if noodle first and bean second is considered different from bean, then noodle, for example). On the other hand, if order is unimportant, then there are 6 ways to choose 2 cans of soup from the 4, as illustrated in the second diagram on the next slide.

  39. Distinctions Between Permutations and Combinations

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