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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## permutation tagged by: Brent@GMATPrepNow ##### This topic has 3 expert replies and 4 member replies ## permutation A committee of 3 people is to be chosen from four married couples. what is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee? 16 24 26 30 32 Pls explain the answer. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12833 messages Followed by: 1247 members Upvotes: 5254 GMAT Score: 770 Manjareev wrote: A committee of 3 people is to be chosen from four married couples. what is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee? 16 24 26 30 32 Pls explain the answer. Here's one approach. Take the task of selecting the 3 committee members and break it into stages. Stage 1: Select the 3 couples from which we will select 1 spouse each. There are 4 couples, and we must select 3 of them. Since the order in which we select the 3 couples does not matter, this stage can be accomplished in 4C3 ways (4 ways) Stage 2: Take one of the 3 selected couples and choose 1 person to be on the committee. There are 2 people in the couple, so this stage can be accomplished in 2 ways. Stage 3: Take one of the 3 selected couples and choose 1 person to be on the committee. There are 2 people in the couple, so this stage can be accomplished in 2 ways. Stage 4: Take one of the 3 selected couples and choose 1 person to be on the committee. There are 2 people in the couple, so this stage can be accomplished in 2 ways. By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus create a 3-person committee) in (4)(2)(2)(2) ways (= 32 ways) Answer = E Cheers, Brent Aside: For more information about the FCP, we have a free video on the subject: http://www.gmatprepnow.com/module/gmat-counting?id=775 _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use our video course along with Sign up for our free Question of the Day emails And check out all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! Junior | Next Rank: 30 Posts Joined 08 Apr 2012 Posted: 18 messages Thank you so much Sir for ur reply. ur explanation is clear. i was just confused for the stage 1 which u had mentioned in the answer. now i know my error of reasoning. thanks. Master | Next Rank: 500 Posts Joined 08 Dec 2010 Posted: 425 messages Followed by: 7 members Upvotes: 56 GMAT Score: 690 four married couples=4*2= 8 people we need to get 6 ppl(excluding 2 married) out of 8 8!/6!/2!=32 _________________ Happy are those who dream dreams and are ready to pay the price to make them come true.(c) In order to succeed, your desire for success should be greater than your fear of failure.(c) Junior | Next Rank: 30 Posts Joined 27 Feb 2015 Posted: 25 messages Upvotes: 1 HI brent, 1st member is selected in 8ways 2nd member in 6 ways (we can't select the spouse) 3rd Member in 4 ways 8*6*4 The order in which the is not important so we must divide by the factorial of the number of choices to eliminate over-counting: (8*6*4)/3! = 8*4 = 32 ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12833 messages Followed by: 1247 members Upvotes: 5254 GMAT Score: 770 Kaustubhk wrote: HI brent, 1st member is selected in 8ways 2nd member in 6 ways (we can't select the spouse) 3rd Member in 4 ways 8*6*4 The order in which the is not important so we must divide by the factorial of the number of choices to eliminate over-counting: (8*6*4)/3! = 8*4 = 32 Perfect! Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use our video course along with Sign up for our free Question of the Day emails And check out all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2444 messages Followed by: 18 members Upvotes: 43 Manjareev wrote: A committee of 3 people is to be chosen from four married couples. what is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee? 16 24 26 30 32 We are given that there are four married couples (or 8 people) and we need to determine the number of ways of choosing 3 people in which no two people are a married couple. This is a special combination problem. Before we tackle this problem, letâ€™s review a combination problem with no restrictions. With no restrictions, the number of ways of choosing 3 people from 8 people is 8C3, which is calculated as follows: 8C3 = 8!/[3!(8-3)!] = 8!/[3!5!] = (8 x 7 x 6)/3! = 56 8, 7, and 6, in the numerator, represent the number of ways the first, second, and third person can be chosen, respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen. However, in this (special combination) problem, 3 people are chosen in which no married couple can serve together on the committee. The first person could be any one of the 8 people. However, once a person is selected, that personâ€™s spouse cannot also be selected for the committee. This reduces the choices for the second person to 6 possible people (one person has already been selected and that personâ€™s spouse now cannot be selected). Once the second person is chosen for the committee, that personâ€™s spouse cannot be chosen. This reduces the number of people who could be chosen as the third person to 4. Therefore, the number of ways of choosing these 3 people is: (8 x 6 x 4)/3! = 32 Thus, there are 32 ways to choose such a committee. Answer: E _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews ### Top Member Master | Next Rank: 500 Posts Joined 15 Oct 2009 Posted: 328 messages Upvotes: 27 Scott@TargetTestPrep wrote: Manjareev wrote: A committee of 3 people is to be chosen from four married couples. what is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee? 16 24 26 30 32 With no restrictions, the number of ways of choosing 3 people from 8 people is 8C3, which is calculated as follows: 8C3 = 8!/[3!(8-3)!] = 8!/[3!5!] = (8 x 7 x 6)/3! = 56 8, 7, and 6, in the numerator, represent the number of ways the first, second, and third person can be chosen, respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen. Answer: E Another view, starting with the above is to now recognize that without the restriction in place, there are among the 56 groups of 3 people a number of groups where a couple is represented and the third person. These need to be subtracted from 56 to comply with the restriction. So how many couples are there ? 4. After accounting for the 2 people in the couple, there are now 8-2 = 6 people left to be the third person. So there are 4x 6 =24 groups of three where a couple is represented. So the correct overall number is then 56-24 = 32 • Award-winning private GMAT tutoring Register now and save up to$200

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