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Learning Activities titled “A Formula for Permutations” and “A Formula for Combinations.” (below) Explain how the Fundamental Counting Principle is used each time the Permutation Formula (nPr) or the Combination Formula (nCr) is applied. Would these be considered independent or dependent events? Or, is it inappropriate to be concerned about whether it is independent/dependent? Explain your thinking on the question.

Learning Activities titled “A Formula for Permutations” and “A Formula for Combinations.” (below) Explain how the Fundamental Counting Principle is used each time the Permutation Formula (nPr) or the Combination Formula (nCr) is applied. Would these be considered independent or dependent events? Or, is it inappropriate to be concerned about whether it is independent/dependent? Explain your thinking on the question.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Learning Activities titled “A Formula for Permutations” and “A Formula for Combinations.” (below) Explain how the Fundamental Counting Principle is used each time the Permutation Formula (nPr) or the Combination Formula (nCr) is applied. Would these be considered independent or dependent events? Or, is it inappropriate to be concerned about whether it is independent/dependent? Explain your thinking on the question.

Consider another problem of this type: An 11-member board of trustees might consider forming a three-person subcommittee. How many possible committees are there? 43 If we solve this using the method we normally use to solve permutations, we would calculate 11.10-9, but this would count a committee consisting of Margaret, Harlow, and Nirit as a different committee than Harlow, Margaret, and Nirit, even though it is the same committee regardless of who is listed first. To solve this problem, we need to divide the total from 11-10-9 by the number of ways we can arrange three people. For example, Margaret, Harlow, and Nirit can be arranged in the following ways: MNH MHN NHM NMH HNM HMN Remember, that we can do this in a much quicker way by thinking about how many choices we have for each spot:3 P₁-3-2-1-6 possibilities. Obviously, all six of these groups are the same exact committee, so they only need to be counted once. This would happen with any subset of three committee members chosen from the 11. One way to eliminate these repetitions is to divide by the number of ways you can arrange three members. In other words: 11-10-9 3-2-1 Note. The material in this section was written by John Redden. Copyright 2014 Flat World Knowledge, Inc.
Transcribed Image Text:Consider another problem of this type: An 11-member board of trustees might consider forming a three-person subcommittee. How many possible committees are there? 43 If we solve this using the method we normally use to solve permutations, we would calculate 11.10-9, but this would count a committee consisting of Margaret, Harlow, and Nirit as a different committee than Harlow, Margaret, and Nirit, even though it is the same committee regardless of who is listed first. To solve this problem, we need to divide the total from 11-10-9 by the number of ways we can arrange three people. For example, Margaret, Harlow, and Nirit can be arranged in the following ways: MNH MHN NHM NMH HNM HMN Remember, that we can do this in a much quicker way by thinking about how many choices we have for each spot:3 P₁-3-2-1-6 possibilities. Obviously, all six of these groups are the same exact committee, so they only need to be counted once. This would happen with any subset of three committee members chosen from the 11. One way to eliminate these repetitions is to divide by the number of ways you can arrange three members. In other words: 11-10-9 3-2-1 Note. The material in this section was written by John Redden. Copyright 2014 Flat World Knowledge, Inc.
Max needs to figure out how many possible route options he has going from one city to another on his two-week vacation in Europe. He has 10 cities on his itinerary and would like to visit at least three of them. How many possible route options does Max have, if he goes to three of the 10 cities during his vacation? Now suppose there are 10 locations and only three stops can be made over a given time period. In this case, how many ordered three-stop routes out of 10 possible stops can be considered? We could use the method: 10-9-8, or we can use the permutation rule where k items are to be selected from n available items. When order matters, this idea is notated where \(_n P_k= \frac{n!}{[(n-k)!}\) To solve this particular routing problem, use n = 10 and k = 3. 10.P 10! (10-3)! 10-9-8-7 X = 10.9.8 720 That is, out of 10 locations, there are 720 possible three-location routes. Note: A permutation function is included on most modern scientific calculators. You may find it in the menu of functions under "probability." Moreover, you can see that this does conform to the fundamental counting principle. There are 10 locations to choose from for the first stop, then nine for the second, and eight for the third.
Transcribed Image Text:Max needs to figure out how many possible route options he has going from one city to another on his two-week vacation in Europe. He has 10 cities on his itinerary and would like to visit at least three of them. How many possible route options does Max have, if he goes to three of the 10 cities during his vacation? Now suppose there are 10 locations and only three stops can be made over a given time period. In this case, how many ordered three-stop routes out of 10 possible stops can be considered? We could use the method: 10-9-8, or we can use the permutation rule where k items are to be selected from n available items. When order matters, this idea is notated where \(_n P_k= \frac{n!}{[(n-k)!}\) To solve this particular routing problem, use n = 10 and k = 3. 10.P 10! (10-3)! 10-9-8-7 X = 10.9.8 720 That is, out of 10 locations, there are 720 possible three-location routes. Note: A permutation function is included on most modern scientific calculators. You may find it in the menu of functions under "probability." Moreover, you can see that this does conform to the fundamental counting principle. There are 10 locations to choose from for the first stop, then nine for the second, and eight for the third.
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