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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 ## Difficult Prob Question ##### This topic has 1 expert reply and 5 member replies ## Difficult Prob Question A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? 28 32 48 60 120 OA 32 Master | Next Rank: 500 Posts Joined 11 Jun 2011 Posted: 370 messages Followed by: 2 members Upvotes: 27 Test Date: Late November Target GMAT Score: 700+ The answer is 32. You can consider simply counting with any parent driving and then adjusting the other 4 family members in 16 ways then multiply the resultant by 2 (which is for the second parent driving). Or you can consider all ways without any restrictions then subtracting the possibilities with the sisters sitting only together. Frong driver seat = 2 ways Front passenger seat= 4 ways Three back passenger seats in 3 * 2 * 1 ways. Total = 2*4*3*2*1 = 48 ways._____(1) Two sisters can sit together only in the back seat: The arrsngements will now be: Front driver seat = 2 ways. Front passenger seat = 2 ways Back seat can be filled by considering both daughters as 1 unit = 2c1 ways = 2 ways As both daughters can interchange their seats = 2ways. Total = 2*2*2*2 = 16 ways.___(2) Number of ways with restrictions = 48-16 = 32 ways.____(1)-(2) Newbie | Next Rank: 10 Posts Joined 14 Jul 2011 Posted: 8 messages Followed by: 1 members Upvotes: 1 winniethepooh wrote: The answer is 32. You can consider simply counting with any parent driving and then adjusting the other 4 family members in 16 ways then multiply the resultant by 2 (which is for the second parent driving). Or you can consider all ways without any restrictions then subtracting the possibilities with the sisters sitting only together. Frong driver seat = 2 ways Front passenger seat= 4 ways Three back passenger seats in 3 * 2 * 1 ways. Total = 2*4*3*2*1 = 48 ways._____(1) Two sisters can sit together only in the back seat: The arrsngements will now be: Front driver seat = 2 ways. Front passenger seat = 2 ways Back seat can be filled by considering both daughters as 1 unit = 2c1 ways = 2 ways As both daughters can interchange their seats = 2ways. Total = 2*2*2*2 = 16 ways.___(2) Number of ways with restrictions = 48-16 = 32 ways.____(1)-(2) Interesting, I did something similiar Who can drive: 2 (mom or dad) Who can ride shotgun: 4 (whoever isn't driving) Than I labeled backseats A B C and gave A and C the value of the sisters. AB, AC, BA, BC, CA, CB, and than eliminated AC + CA since they can't sit next to each other leaving me with 4 possible combinations. 2*4*4 = 32 Just wondering, how did you get Front driver and Front Passenger to be both 2? Master | Next Rank: 500 Posts Joined 11 Jun 2011 Posted: 370 messages Followed by: 2 members Upvotes: 27 Test Date: Late November Target GMAT Score: 700+ Front Driver = Mom or dad Front Passenger: = Son or Father(This is to satisfy the condition that both sisters are sitting together, and theycan sit together only at the back seats). ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15383 messages Followed by: 1872 members Upvotes: 13060 GMAT Score: 790 chaitanya.mehrotra wrote: A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? 28 32 48 60 120 OA 32 An alternate approach: One daughter in the front passenger seat: Since one of the 2 parents must drive, number of choices for the driver's seat = 2. Since either of the 2 daughters can be in the front passenger seat, number of choices = 2. Number of ways to arrange the 3 remaining people = 3! = 6. Multiplying the results above, we get: Number of arrangements = 2*2*6 = 24. The two daughters separated in the back seat: Since one of the 2 parents must drive, number of choices for the driver's seat = 2. Since either of the 2 daughters can be in the leftmost back seat, number of choices = 2. Since the 1 remaining daughter must be in the rightmost back seat, number of choices = 1. Number of ways to arrange the 2 remaining people = 2! = 2. Multiplying the results above, we get: Number of arrangements = 2*2*1*2 = 8. Thus, total arrangements = 24+8 = 32. The correct answer is B. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. Master | Next Rank: 500 Posts Joined 11 Jun 2011 Posted: 370 messages Followed by: 2 members Upvotes: 27 Test Date: Late November Target GMAT Score: 700+ Mitch can we solve it by the normal way, that is considering 4 options in the front passenger seat and then considering the restrictions at the back seat. I understand what you've done is similar, but can we conclude it without having to add up two individual answers? Senior | Next Rank: 100 Posts Joined 13 Jun 2010 Posted: 44 messages Followed by: 3 members Upvotes: 5 Test Date: 9 July 2013 Target GMAT Score: 730 GMAT Score: 720 GMATGuruNY wrote: An alternate approach: One daughter in the front passenger seat: Since one of the 2 parents must drive, number of choices for the driver's seat = 2. Since either of the 2 daughters can be in the front passenger seat, number of choices = 2. Number of ways to arrange the 3 remaining people = 3! = 6. Multiplying the results above, we get: Number of arrangements = 2*2*6 = 24. The two daughters separated in the back seat: Since one of the 2 parents must drive, number of choices for the driver's seat = 2. Since either of the 2 daughters can be in the leftmost back seat, number of choices = 2. Since the 1 remaining daughter must be in the rightmost back seat, number of choices = 1. Number of ways to arrange the 2 remaining people = 2! = 2. Multiplying the results above, we get: Number of arrangements = 2*2*1*2 = 8. Thus, total arrangements = 24+8 = 32. The correct answer is B. I get confused when to add or multiple, like for the final step in this question. Can someone please shed some light. Thank you • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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