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 considerforming a three-person subcommittee. How many possible committees are there?43If we solve this using the method we normally use to solve permutations, we wouldcalculate 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 thesame 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 wayswe can arrange three people. For example, Margaret, Harlow, and Nirit can be arrangedin the following ways:MNH MHN NHM NMH HNM HMNRemember, that we can do this in a much quicker way by thinking about how manychoices 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 becounted once. This would happen with any subset of three committee members chosenfrom the 11.One way to eliminate these repetitions is to divide by the number of ways you can arrangethree members. In other words:11-10-93-2-1Note. The material in this section was written by John Redden. Copyright 2014 Flat WorldKnowledge, Inc. Max needs to figure out how many possible route options he has going from one city toanother on his two-week vacation in Europe. He has 10 cities on his itinerary and wouldlike to visit at least three of them. How many possible route options does Max have, if hegoes 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 timeperiod. In this case, how many ordered three-stop routes out of 10 possible stops can beconsidered? We could use the method: 10-9-8, or we can use the permutation rulewhere k items are to be selected from n available items. When order matters, this idea isnotated where$_n P_k= \frac{n!}{[(n-k)!}$To solve this particular routing problem, use n = 10 and k = 3.10.P10!(10-3)!10-9-8-7X= 10.9.8720That 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, andeight for the third.

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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 considerforming a three-person subcommittee. How many possible committees are there?43If we solve this using the method we normally use to solve permutations, we wouldcalculate 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 thesame 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 wayswe can arrange three people. For example, Margaret, Harlow, and Nirit can be arrangedin the following ways:MNH MHN NHM NMH HNM HMNRemember, that we can do this in a much quicker way by thinking about how manychoices 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 becounted once. This would happen with any subset of three committee members chosenfrom the 11.One way to eliminate these repetitions is to divide by the number of ways you can arrangethree members. In other words:11-10-93-2-1Note. The material in this section was written by John Redden. Copyright 2014 Flat WorldKnowledge, Inc. Max needs to figure out how many possible route options he has going from one city toanother on his two-week vacation in Europe. He has 10 cities on his itinerary and wouldlike to visit at least three of them. How many possible route options does Max have, if hegoes 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 timeperiod. In this case, how many ordered three-stop routes out of 10 possible stops can beconsidered? We could use the method: 10-9-8, or we can use the permutation rulewhere k items are to be selected from n available items. When order matters, this idea isnotated where$_n P_k= \frac{n!}{[(n-k)!}$To solve this particular routing problem, use n = 10 and k = 3.10.P10!(10-3)!10-9-8-7X= 10.9.8720That 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 &#34;probability.&#34;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, andeight for the third.

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