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07302010, 07:36 PM  #1 
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GMAT Quant Questions Help
It's a permutation/combination problem. I can get most of these correct, but for some reason this one is a PITA.
A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that both of them will sit next to only one other student from the group? The Answer is 10% First, I assume you need to find out the total number of of seating arrangements. It's 5!=120. Then, I find out how many options are available so that they sit next to only 1 other student. 5!/(3!2!)=10... Obviously either this number or the first are incorrect, since 10/120 is not 10%. So, I tried to figure out all the possible ways that could sit, let L be Lisa, B be Bob, and x be the other students. LBxxx BLxxx xBLxx xLBxx xxLBx xxBLx xxxBL xxxLB BxxxL LxxxB That is still 10. I cant quite figure out what I'm doing wrong. I think it's probably the way I'm reading the question, so hopefully some fresh eyes can help.
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07302010, 07:43 PM  #2 
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does 1 2 7 and 8 count since only one of them is sitting next to one other person from the group, or does sitting next to each other count?
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07302010, 08:08 PM  #3  
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5!/2!=60, but not sure the reasoning for that...
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07302010, 08:30 PM  #4 
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Just curious...where did you get the question from? I'm asking because I'm studying for the GMAT as well, taking it on August 11th. I'm using the prep book from the folks at GMAT to study and the questions come with explanations.

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07302010, 08:52 PM  #7 
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why would that number matter it says they are all sitting next to each other.
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07302010, 08:59 PM  #8  
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bloxx lboxx xxolb xxobl In each of these cases only one sits next to another person, but not both. In every other conceivable case, both of them individually must sit next to at least one another person. loxob boxol oblox olbox xolbo xoblo obolo olobo etc... Last edited by radix; 07302010 at 09:29 PM. 

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07302010, 09:11 PM  #9  
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07302010, 09:28 PM  #10 
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07302010, 09:44 PM  #12 
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OK.... I got it, I just didn't interpret "one other person" to mean "one other specific person". When you read it that way, everything changes.
Let's call this one person Frank. The problem should have been more specific IMO, but for the sake of solving, I'll call him Frank. The probability that Frank takes a seat that is not on one of the ends is 3:5, or .60. The probability that the next person, say Bob, sits next to Frank is 2:4 (he can sit on either side) or .50. The probability that Linda takes the other seat next to Frank is 1:3, or .33 repeating. Therefore: Code:
$ perl le 'print 0.6 * 0.5 * 0.333333333' 0.0999999999 So the answer is 10%. 
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07302010, 10:01 PM  #14 
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If Frank sat on the end, then Bob and Linda could not both sit next to him. The problem is badly worded, but in order to both sit next to this one person (hypothetically Frank), they must of course flank him or her on either side. This is not possible if this one person (Frank) sits in an end seat. The mistake the OP made is thinking that this problem should be solved using permutations. It shouldn't. It should be solved my multiplying the probabilities of successive events.

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07302010, 10:15 PM  #15 
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Are you sure that both Frank and Bob have to sit next to only one person (Linda) or do they individually have to sit next to other person.. like BOXOF or XOBFO. The question is veryy badly worded. So how would you get .90 for possibilities they wont only sit next to one person?
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07302010, 10:21 PM  #16  
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They have to sit like this: xxbfl xxlfb xlfbx xbflx bflxx lfbxx In order for both Bob and Linda to sit next to this person (Frank), he must be in between them, otherwise only one of them is sitting next to him. 

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07302010, 10:32 PM  #17 
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what my question was is that do both of them have to sit next to the same person or can they each sit next to 1 other person like LSxFB where both Linda and Bob are sitting next to one person
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07302010, 10:38 PM  #18  
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2 Linda is sitting next to 2 people 3 Linda and bob are sitting next to 2 people same with 4 5 Linda is next to 2 people 6 Bob is next to 2 people
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07302010, 10:45 PM  #19 
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You're still misunderstanding the problem. It asks "what is the probability that both of them will sit next to only one other student from the group?". The only way they can both sit next to only one other student from the group, is if that student sits between them.

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07302010, 10:52 PM  #20 
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slightly hijacking this thread temporarily, but i just took the gmat about a month ago, and i have to vouch for the manhattan gmat books. those books are amazing! helped me a lot and explained a lot of the little minute details and fundamentals that i never really understood 100%. definitely better than the official gmat books, and i took the kaplan class prob 23 years ago, and def blows that out of the water.
highly recommended
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07302010, 11:17 PM  #21  
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07302010, 11:24 PM  #22  
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Consider the problem this way. Instead of: Quote:
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Then the only possible positions that work with respect to Bob, Lisa, and Frank are: jgbfl jglfb jlfbg jbflg bfljg lfbjg The positions of Jen and George don't matter with respect to probability. I don't know how else to explain it, maybe your teacher can do a better job. Last edited by radix; 07302010 at 11:30 PM. 

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