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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 ## Counting Problem - Company recruitment tagged by: Brent@GMATPrepNow ##### This topic has 1 expert reply and 9 member replies ## Counting Problem - Company recruitment A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location? A - 180 B - 325 C - 504 D - 730 E - 1260 OA is c Master | Next Rank: 500 Posts Joined 01 May 2013 Posted: 149 messages Followed by: 9 members Upvotes: 54 Top Reply rkiran9 wrote: ...I followed grid approach, i.e out of 9 people 3 needs to be selected means 6 are not selected.. so 9!/3!6!.... but its wrong.... It is not just selecting 3 people out of 9 but after selecting you have to send the selected people in 3 different cities. So, you have to consider the possible number of arrangements among them which is 3! So, the correct answer should be 9!/3!6! x 3! = 9!/6! Hope that helps. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 13046 messages Followed by: 1253 members Upvotes: 5254 GMAT Score: 770 faraz_jeddah wrote: A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location? A - 180 B - 325 C - 504 D - 730 E - 1260 OA is c Take the task of filling the 3 positions and break it into stages. Stage 1: Select someone for the LA position There are 9 candidates, so we can complete stage 1 in 9 ways. Stage 2: Select someone for the New York position There are 8 candidates remaining, so we can complete stage 2 in 8 ways. Stage 3: Select someone for the Boston position There are 7 candidates remaining, so we can complete stage 3 in 7 ways. By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus fill the 3 positions) in (9)(8)(7) ways (= 504 ways) Answer: C 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 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! Master | Next Rank: 500 Posts Joined 18 Apr 2013 Posted: 358 messages Followed by: 7 members Upvotes: 42 Test Date: 17-Nov Target GMAT Score: 690 GMAT Score: 730 Thanks Brent! Just realized I was making a multiplication error. I was getting worked up because the problem looked straight forward. Junior | Next Rank: 30 Posts Joined 11 Apr 2013 Posted: 12 messages Upvotes: 1 Brent@GMATPrepNow wrote: faraz_jeddah wrote: A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location? A - 180 B - 325 C - 504 D - 730 E - 1260 OA is c Take the task of filling the 3 positions and break it into stages. Stage 1: Select someone for the LA position There are 9 candidates, so we can complete stage 1 in 9 ways. Stage 2: Select someone for the New York position There are 8 candidates remaining, so we can complete stage 2 in 8 ways. Stage 3: Select someone for the Boston position There are 7 candidates remaining, so we can complete stage 3 in 7 ways. By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus fill the 3 positions) in (9)(8)(7) ways (= 504 ways) Answer: C 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 I knw something is wrong with my approach but I wanna understand y? Select 3 guys out of 9 -- 9C3 .. And then send them to any of 3 locations . 3C1 so answer is 9C3* 3C1 = 252.. Why is this wrong approach? Pls explain .. will be helpful Senior | Next Rank: 100 Posts Joined 04 Jan 2013 Posted: 38 messages Upvotes: 11 Quote: I knw something is wrong with my approach but I wanna understand y? Select 3 guys out of 9 -- 9C3 .. And then send them to any of 3 locations . 3C1 so answer is 9C3* 3C1 = 252.. Why is this wrong approach? Pls explain .. will be helpful Sending them to any of the three locations wont be 3C1. Sending A, B, C to locations P, Q, R. We aren't choosing 1 out of 3, but choosing 1 out of 3 and then choosing another 1 out of remaining 2., that'll be 3C1*2C1 = 6. Newbie | Next Rank: 10 Posts Joined 05 Nov 2012 Posted: 5 messages rac.nishu wrote: Brent@GMATPrepNow wrote: faraz_jeddah wrote: A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location? A - 180 B - 325 C - 504 D - 730 E - 1260 OA is c Take the task of filling the 3 positions and break it into stages. Stage 1: Select someone for the LA position There are 9 candidates, so we can complete stage 1 in 9 ways. Stage 2: Select someone for the New York position There are 8 candidates remaining, so we can complete stage 2 in 8 ways. Stage 3: Select someone for the Boston position There are 7 candidates remaining, so we can complete stage 3 in 7 ways. By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus fill the 3 positions) in (9)(8)(7) ways (= 504 ways) Answer: C 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 I knw something is wrong with my approach but I wanna understand y? Select 3 guys out of 9 -- 9C3 .. And then send them to any of 3 locations . 3C1 so answer is 9C3* 3C1 = 252.. Why is this wrong approach? Pls explain .. will be helpful Hi Brent, I followed grid approach, i.e out of 9 people 3 needs to be selected means 6 are not selected.. so 9!/3!6!.... but its wrong.... Regards Ravikiran D Legendary Member Joined 18 Jun 2012 Posted: 512 messages Followed by: 19 members Upvotes: 42 Test Date: October '13 Target GMAT Score: 740 just forget about three chosen people, remember their are 9 eligibale people and 3 vacant positions. first seat at one location can be filled in 9 ways second seat at second location in 8 ways and 3rd seat at third location in 7 ways so the answer will be 9.8.7 = 504 ways However if the question asks to choose three members out of 9 eligible candidates and order isnt important than the answer will be 9.8.7/1.2.3 = 84 different combinations of 3 members. _________________ Work hard in Silence, Let Success make the noise. If you found my Post really helpful, then don't forget to click the Thank/follow me button. Newbie | Next Rank: 10 Posts Joined 23 Jul 2013 Posted: 2 messages Initially lets keep aside the location requirement and just figure out the number of 3 member combinations for 3 vacant posts from the 9 candidates available. it comes to 9C3 = 9!/(9-3)! * 3! = 84. So we have 84 possible 3 candidate combinations. The 3 candidates (Say A, B and C) for each combination can be placed at 3 locations in the following 6 ways: Los Angeles || New York || Boston. 1 . A || B || C 2 . A || C || B 3 . B || A || C 4 . B || C || A 5 . C || A || B 6 . C || B || A i.e 3 candidates can be places in 3 places in 3P3 = 3! /(3-3!) = 6 ways. Hence the total number of ways the company can select 3 candidates for 3 locations is 84*6 = 504 ways. Hope this helps... Master | Next Rank: 500 Posts Joined 20 Sep 2014 Posted: 164 messages Upvotes: 1 Target GMAT Score: 650 A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location? A - 180 B - 325 C - 504 D - 730 E - 1260 LA: 9 ways NY: 8 ways B: 7 ways 9*8*7 = 504 _________________ Hard work brings success! Newbie | Next Rank: 10 Posts Joined 15 Jul 2015 Posted: 6 messages Ravikiran the issue about your approach is that you missed the second step, The first step of 9C3 is correct but the next step was to use the FCP for unique Objects which we had three options in step 2 I.e 3x2x1. So we have step one multiplied by step 2. 9C3 x 3x2x1= 84x6=504 • FREE GMAT Exam Know how you'd score today for$0

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