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Simple probability question. Need some clarity on concepts [#permalink]
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21 May 2009, 10:19
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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?? == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: Simple probability question. Need some clarity on concepts [#permalink]
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23 May 2009, 23:52
If you name the couples A and B, you end up with five people (AM AF SP BM BF), if you look at the 5 chairs as 5 spaces on each row, you can calculate how many possibilities there are that the couples are seated together, and not touching each other. The single person (SP) must be in the middle for the couples not to be seated next to eachother. 1 2 3 4 5 AM AF SP BM BF AF AM SP BM BF AM AF SP BF BM AF AM SP BF BM BM BF SP AM AF BF BM SP AM AF BM BF SP AF AM BF BM SP AF AM
8 possible outcomes of the seating that resulting in the stimulus being met, out of the possible outcomes (5!, or 5*4*3*2*1)=120, so the probability of the stimulus being met is 8/120, or .66%. Answer choices would be helpful to...



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Re: Simple probability question. Need some clarity on concepts [#permalink]
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24 May 2009, 01:10
dk94588 +1 for the thorough explanation!



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Re: Simple probability question. Need some clarity on concepts [#permalink]
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24 May 2009, 04:18
I am afraid. The OA says 2/5....Not able to understand the explanation clearly.. can anyone explain alternate method...
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



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Re: Simple probability question. Need some clarity on concepts [#permalink]
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24 May 2009, 07:56
Yeah I took for the granted that it was random seating and that the couples were separate people. Again, answer choices would help you decide which method to use. Where is this from? because the question itself seems a little unreliable..



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Re: Simple probability question. Need some clarity on concepts [#permalink]
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24 May 2009, 08:20
Good question + good explanation! My way of solving was muuuch slower. Possible answer choices are A] 1/5 B] 2/5 C] 3/5 D] 4/5 E] 1/20 People: couple A: A1 and A2; couple B: B1 and B2; single person: X Places: 1 2 3 4 5 (1) Ways of couple A sitting together: 4 ways of placing the couple (A345, 1A45, 12A5, 123A), 3! ways of placing 3 other people in 3 possible places, 2 ways of arranging of people inside the couple (A1A2 or A2A1). Total number of ways: 4 x 3! x 2 (the same as 2 x 4!) (2) The same for couple B: total number of ways: 4 x 3! x 2 (the same as 2 x 4!) But (1) and (2) both include the number of ways in which both couples can sit together (e.g. in (1) there is a variant A1 A2 3 B1 B2, as well as in (2)), it means that if we sum up (1) and (2), we will have: number of ways only A sits together + number of ways both sit together + number of ways only B sits together + number of ways both sit together (Venn diagram). Thus, we have to subtract one set of both sitting together. (3) Both couples sitting together: 3 ways of placing X (X _ _ _ _, _ _ X _ _ or _ _ _ _ X), 2 ways of placing couples (A on the left and B on the right of vice versa), 2 ways of arranging people inside couple A (A1A2 and A2A1), 2 ways of arranging people inside couple B, total number of ways: 3 x 2 x 2 x 2 (the same as 2*2*3! = 4!) Then the total number of ways at least one couple sits together: (1) + (2)  (3) = 72, possibility is 72/120 = 3/5, possibility of NOT sitting together: 1  3/5 = 2/5
As alternate method I can write down my first attempt to solve this problem, but my method was very time consuming..



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Re: Simple probability question. Need some clarity on concepts [#permalink]
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24 May 2009, 08:37
sorry to keep posting irrelevant stuff but I got it now. I took it to mean that the couples always sat next to each other and that the couples themselves could not sit next to each other. If you mean to say that the couples themselves do not sit next to each other, regardless of where they are in relation to the other people, then yes 2/5 is the correct answer. misunderstood the stimulus.



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Re: Simple probability question. Need some clarity on concepts [#permalink]
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25 May 2009, 23:00
Is it possible to solve this kind of question in 2 min. ??? I am asking because many questions in this forum takes me more then 2 min. to solve. Is it just me or the questions are really tough ?



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Re: Simple probability question. Need some clarity on concepts [#permalink]
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27 May 2009, 14:24
the ones in the forum are usually the hardest ones the posters can find. kind of a mind tease. Or its the ones that they themselves couldn't figure out, so that would make them hard as well. You have to do the easy ones the fastest you can to make up for the time it takes for the hard ones...



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Re: Simple probability question. Need some clarity on concepts [#permalink]
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28 May 2009, 08:08
bandit wrote: 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?? Can you make it clear that how the chairs are arranged? Circular or in a row? I think the probability is different for each scenario. mdfrahim wrote: Is it possible to solve this kind of question in 2 min. ??? I am asking because many questions in this forum takes me more then 2 min. to solve. Is it just me or the questions are really tough ? IMO, the 2 minuet issue is for all not for a single question. Some question can be solved in few seconds, 5 or 10, some might take more than 2 minuets. In average, yes 2 minuets is standard for a question.
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Re: Simple probability question. Need some clarity on concepts [#permalink]
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28 May 2009, 14:12
I am pretty sure the assumption is that they are in a straight line, they would have said they were in a circle if they were. The GMAT people wouldn't mean for you to assume they were in a circle in a question like that. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.




Re: Simple probability question. Need some clarity on concepts
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28 May 2009, 14:12






