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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 ## If there are 4 pairs of twins, and a committee will be ##### This topic has 2 expert replies and 0 member replies ### Top Member ## If there are 4 pairs of twins, and a committee will be ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group? A. 32 B. 24 C. 56 D. 44 E. 40 OA A Source: GMAT Prep ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12827 messages Followed by: 1247 members Upvotes: 5254 GMAT Score: 770 BTGmoderatorDC wrote: If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group? A. 32 B. 24 C. 56 D. 44 E. 40 OA A Source: GMAT Prep Here's one approach. Take the task of selecting the 3 committee members and break it into stages. Stage 1: Select the 3 twins from which we will select 1 spouse each. There are 4 sets of twins, and we must select 3 of them. Since the order in which we select the 3 twins does not matter, this stage can be accomplished in 4C3 ways (4 ways) If anyone is interested, I have a video on calculating combinations (like 4C3) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789 Stage 2: Take one of the 3 selected pairs of twins and choose 1 person to be on the committee. There are 2 people, so this stage can be accomplished in 2 ways. Stage 3: Take one of the 3 selected pairs of twins and choose 1 person to be on the committee. There are 2 people, so this stage can be accomplished in 2 ways. Stage 4: Take one of the 3 selected pairs of twins and choose 1 person to be on the committee. There are 2 people, 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: A Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting/video/775 You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat-counting/video/776 Then you can try solving the following questions: EASY - http://www.beatthegmat.com/what-should-be-the-answer-t267256.html - http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html - http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html - http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html - http://www.beatthegmat.com/simple-one-t270061.html MEDIUM - http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html - http://www.beatthegmat.com/arabian-horses-good-one-t150703.html - http://www.beatthegmat.com/sub-sets-probability-t273337.html - http://www.beatthegmat.com/combinatorics-problem-t273180.html - http://www.beatthegmat.com/digits-numbers-t270127.html - http://www.beatthegmat.com/doubt-on-separator-method-t271047.html - http://www.beatthegmat.com/combinatorics-problem-t267079.html DIFFICULT - http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html - http://www.beatthegmat.com/permutation-and-combination-t273915.html - http://www.beatthegmat.com/permutation-t122873.html - http://www.beatthegmat.com/no-two-ladies-sit-together-t275661.html - http://www.beatthegmat.com/combinations-t123249.html 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: 2424 messages Followed by: 18 members Upvotes: 43 BTGmoderatorDC wrote: If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group? A. 32 B. 24 C. 56 D. 44 E. 40 OA A Source: GMAT Prep The number of ways to select the 3 pairs from 4 pairs is 4C3 = 4. Since there can be no siblings on the board each twin can be selected in 2C1 ways, so: 2C1 x 2C1 x 2C1 = 2 x 2 x 2 = 8 So the total number of ways to select the committee is 4 x 8= 32. Alternate Solution: For the first member, there are 8 choices. Since the sibling of the first member cannot be chosen, there are 6 choices for the second member. By the same logic, there are 4 choices for the last member. Notice that the 8 x 6 x 4 choices count each committee 3! times; therefore, there are (8 x 6 x 4)/3! = 8 x 4 = 32 possible committees. Answer: A _________________ 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. 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