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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 Math Problem #15 This topic has 3 member replies Difficult Math Problem #15 OA coming after some people attempt an answer: TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above? Senior | Next Rank: 100 Posts Joined 03 Sep 2006 Posted: 34 messages Total possibilites: 5! if i arrange couple A with single B like this: AAB then i can put couple C in the places marked by '_' : _A_A_B_ so ill have 4P2 = 12 options (A is man and female i assume so thats why it is 4P2) i can make 6 like this, so i have 72 options so total there are 72 options to arrange couples to not sit by each other out of 5! 72/120 = 3/5 i hope im right Newbie | Next Rank: 10 Posts Joined 17 Sep 2006 Posted: 3 messages let AA BB C stand for 5 people if C sit on the first chair, 4 choices for the second chair and 2 choice for the third chair = 4*2= 8. Same thing happens if C sit on the fifth chair = 8 Now, if C sit on the second chair, 4 choices (for the first chair) * 2 (for the third) = 8. Similarly, if C sit on the fourth chair, 4*2 = 8 finally, if C in the middle (third chair) 4 choices for the first*2 choices for the second*2 choices for the fourth = 16 Total = 8+8+8+8+16 = 48 Probability = 48/5! = 2/5 = 40% Hope right Master | Next Rank: 500 Posts Joined 27 Jun 2006 Posted: 354 messages Followed by: 5 members Upvotes: 11 here's the OA: Ways in which the first couple can sit together = 2*4! (1 couple is considered one unit) Ways for second couple = 2*4! These cases include an extra case of both couples sitting together Ways in which both couple are seated together = 2*2*3! = 4! (2 couples considered as 2 units- so each couple can be arrange between themselves in 2 ways and the 3 units in 3! Ways) Thus total ways in which at least one couple is seated together = 2*4! + 2*4! - 4! = 3*4! Total ways to arrange the 5 ppl = 5! Thus, prob of at least one couple seated together = 3*4! / 5! = 3/5 Thus prob of none seated together = 1 - 3/5 = 2/5 • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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