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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 ## What is the difference between the number of three-member tagged by: AAPL ##### This topic has 4 expert replies and 0 member replies ### Top Member ## What is the difference between the number of three-member ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Princeton Review What is the difference between the number of three-member committees that can be formed from a group of nine members and the total number of ways there are to arrange the members of such a committee? A. 0 B. 84 C. 252 D. 420 E. 504 OA D. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12921 messages Followed by: 1248 members Upvotes: 5254 GMAT Score: 770 AAPL wrote: Princeton Review What is the difference between the number of three-member committees that can be formed from a group of nine members and the total number of ways there are to arrange the members of such a committee? A. 0 B. 84 C. 252 D. 420 E. 504 The number of three-member committees that can be formed from a group of nine members Since the order in which we select the 3 people does not matter, we can use combinations. We can choose 3 people from 9 people in 9C3 ways 9C3 = (9)(8)(7)/(3)(2)(1) = 84 three-member committees The total number of ways there are to ARRANGE the members of such a committee We can arrange n objects in n! ways So, we can arrange the 3 people (in the committee) in 3! ways (= 6 ways) DIFFERENCE = 84 - 6 = 78 78 is not among the answer choices. hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm! It appears (given the official answer) that we're supposed to arrange the 3 members in each of the 84 three-member committees So, for each of the 84 three-member committees, we can arrange the three people in 6 ways So, the total number of arrangements = (84)(6) = 504 So, the DIFFERENCE = 504 - 84 = 420 (D) IMO, this question is too ambiguous to be GMAT-worthy. Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these 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 Elite Legendary Member Joined 23 Jun 2013 Posted: 10197 messages Followed by: 494 members Upvotes: 2867 GMAT Score: 800 Hi All, To start, I agree with Brent; the wording of this prompt is 'clunky' - and the GMAT writers word their questions in a far more rigorous and specific fashion than what we see here. That having been said, the basic concepts involved here are Combinations and Permutations. We're asked for the difference between the number of three-member committees that can be formed from a group of nine members and the total number of ways there are to arrange the members of such a committee. The intent of this question is to ask for the difference in the number of possible 3-person groups and the number of ways to arrange 3 of the 9 people 'in a row.' For the number of 3-person groups, we can use the Combination Formula: N!/K!(N-K)! = 9!/3!(9-3)! = (9)(8)(7)/(3)(2)(1) = 504/6 = 84 The number of ways to arrange 3 of the 9 people in a row = (9)(8)(7) = 504 The difference is 504 - 84 = 420 Final Answer: D GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 Quote: [Possible alternate question stem to avoid BrentÂ´s and RichÂ´s (and also my) insatisfaction] What is the difference between the number of ways to select a three-member committee from a group of nine members, taking or not taking into account the order of selection of these members? A. 0 B. 84 C. 252 D. 420 E. 504 $$? = 3!C\left( {9,3} \right) - C\left( {9,3} \right)\,\,\mathop = \limits^{\left( * \right)} \,\,\,84\left( {3! - 1} \right) = 5 \cdot 84 = 420$$ $$\left( * \right)\,\,\,C\left( {9,3} \right) = {{9 \cdot 8 \cdot 7} \over {3 \cdot 2}} = 3 \cdot 4 \cdot 7 = 84$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2624 messages Followed by: 18 members Upvotes: 43 AAPL wrote: Princeton Review What is the difference between the number of three-member committees that can be formed from a group of nine members and the total number of ways there are to arrange the members of such a committee? A. 0 B. 84 C. 252 D. 420 E. 504 OA D. The number of 3-member committees that can be formed from 9 people (i.e., order doesnâ€™t matter) is 9C3 = 9!/(3! x 6!) = (9 x 8 x 7)/3! = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84. The number of ways to form the committees and arrange the members (i.e., order matters) is 9P3 = 9!/6! = 9 x 8 x 7 = 504. Thus, the difference is 504 - 84 = 420. Answer: D • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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