How many ways can a committee of 3 people be selected from a group of 10 people?
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However, understanding it, you can calculate the product of three numbers beginning at 5 and counting down: 5 times 4 times 3, and then divide by the product of the three numbers beginning at 3 and counting down: 3 times 2 times 1. Show Final answer is (5*4*3) over (3*2*1) or 60/6 = 10. Answer by CharStar(110) You can put this solution on YOUR website! Now solve When selecting people to serve on committees, we need to be careful whether we use permutations or combinations (or both). Typically, if the committee members are ranked, such as chairman, vice chairman, recorder, we must use permutations. So if we want to know how many ways to select a ranked committee of 3 from 12 people, we would write P(12,3) = 1320 ways If the committee members are not ranked (they are simply members), then we use combinations. For example, the number of ways to choose a committee of three from 12 potential members is C(12,3) = 220 ways On some committees, there are ranked members and unranked members. For instance, if we had 12 people to choose from for a committee of 4 with a chairman and vice chairman, we would think of this using a slot diagram. One slot involves choosing the chairman and vice chairman and the other slot involves choosing the other two members. The number of ways to choose the chairman and vice chairman is P(12,2) = 132 Since the other two members are not ranked, we choose them from the remaining 10 members, C(10,2) = 45 Multiplying these numbers gives P(12,2) ⋅ C(10,2) = 132 ⋅ 45 = 5940 In this case we need to use permutations and combinations since the committee contains ranked and unranked members. Now let’s look at a tricky problem: Problem Twelve people are to be distributed across three committees of four people each. How many ways is there to do this? Solution Let’s think of this with a slot diagram where each slot corresponds to choosing a committee.  When choosing the first committee, we are choosing four people from twelve so the first slot becomes Once we have chosen the first committee, there are eight people top choose from for the second committee: Finally, we choose the third committee from the remaining people: Putting in the correct values for each combination gives This gives a total of 34,650 possible committees. However, we have overlooked something. Using a slot diagram builds in an ordering issue. In counting the committees, we have counted different rearrangements as being different. In other words, this number counts C1, C2, C3 as being different from C3, C2, C1…and also different from C2, C1, C3. If we had assigned each committee a different task, we would want to rearrange the committees among the tasks and count those rearrangements. In our problem, each committee is equal so we do not want the rearrangements to count. Since there are 3 ·2 · 1 = 6 ways to rearrange the committees, the total number of committees is How many combinations of 3 students can be selected from a group of 10 students?10! / 7! Because the order is not significant, we divide this answer by 3!, giving us 720 /6 = 120.
How many ways can 3 people be selected from a group of 11?Answer and Explanation: There are 165 ways of making a group of 3 out of 11 people.
How many ways can 2 people be selected from a group of 10 people?The number of ways to select 2 people out of 10, when the order of selection matters, is computed using the permutation formula as shown below. Thus, there are 90 ways to select 2 people out of 10 when the order of selection matters.
How many ways can a committee of 3 people be selected from a group of 20?The order of the people matters, because a different order results in different people being assigned to the different positions, and thus we should use the definition of permutation \textbf{permutation} permutation. Thus there are 6840 different committees.
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