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Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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20 Nov 2019, 18:26
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Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to each other in 6 adjacent seats in the second row of the theater. If Rob, Ray, and Susan will not sit next to each other, in how many different arrangements can the six people sit? A. 144 B. 576 C. 714 D. 686 E. 696
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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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20 Nov 2019, 18:40
There are 2 ways that come to mind to solve this, but I'm getting different answers. I've been trying to figure this out for a while. Just can't see where I'm going wrong.
Option 1
If 3 of them are to not sit next to each other, those three can sit in the following set up. In the diagram below, Y denotes a seat filled with someone who does not want to sit next to another person, and _ denotes an empty seat which can be filled by those without seating restrictions.
Y _ Y _ Y _ Y _ _ Y _ Y Y _ Y _ _ Y _ Y _ Y _ Y
FOR EACH of the options above, the 3 of them can sit in 3! ways = 6 ways. In addition to that, FOR EACH of the options, the remaining empty seats can also be filled in 3! ways = 6 ways.
Hence, the ways the 3 of them won't sit next to each other is 4*6*6 = 144
Option 2
Calculate total number of seating arrangements  Ways 3 of them can sit together
Total number of seating arrangements = 6! = 720
For ways 3 of them can sit together, pretend the group is a group of 4 individuals instead of 6 (because 3 of them have to sit together anyway). If you had 4 individuals, you'd have 4! ways = 24 ways. However, one of those individuals is actually a group of 3 people, and the ways that group of 3 people can sit is 3! = 6. This 6 multiplied by the 24 calculated earlier gives you the number of ways 3 of them can sit together, which is 144.
Total seating arrangements  Ways of sitting together = 720  144 = 576
I'm getting different answers in Option 1 and Option 2, and I'm not sure if either of them is even correct. Any help would be appreciated. Thanks everyone!



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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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20 Nov 2019, 19:02
Total arrangements  ( arrangement when three of them are together)
6!  ( 4! * 3!) = 576
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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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20 Nov 2019, 22:35
johnd32 wrote: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to each other in 6 adjacent seats in the second row of the theater. If Rob, Ray, and Susan will not sit next to each other, in how many different arrangements can the six people sit? Very simple question total person are 6. 3 person cannot sit with each other 6!( 4!*3!)=576



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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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20 Nov 2019, 22:37
Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to each other in 6 adjacent seats in the second row of the theater. If Rob, Ray, and Susan will not sit next to each other, in how many different arrangements can the six people sit? Total arrangement = Three of them together + Three of them not together 6! = 4! * 3! + Three of them not together (4! = L M J + Three of them, 3! = Rob, Ray, Susan interchangeable) Three of them not together = 720144 = 576.
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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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20 Nov 2019, 23:14
johnd32 wrote: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to each other in 6 adjacent seats in the second row of the theater. If Rob, Ray, and Susan will not sit next to each other, in how many different arrangements can the six people sit?
A. 144 B. 576 C. 714 D. 686 E. 696 Similar questions: https://gmatclub.com/forum/gregmarcia ... 42275.htmlhttps://gmatclub.com/forum/sixfriends ... fl=similarhttps://gmatclub.com/forum/6personsar ... fl=similarhttps://gmatclub.com/forum/7peopleab ... fl=similarCheck other Arrangements in a Row and around a Table questions in our Special Questions Directory.
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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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20 Nov 2019, 23:29
BunuelI believe the question is rather ambiguous. If Rob, Ray, and Susan will not sit next to each other, in how many different arrangements can the six people sit?I guess this should mean that no two of them can sit together in addition to the fact that the three cannot sit together. In this case the answer would be 144. Can you please clarify?



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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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20 Nov 2019, 23:32
firas92 wrote: BunuelI believe the question is rather ambiguous. If Rob, Ray, and Susan will not sit next to each other, in how many different arrangements can the six people sit?I guess this should mean that no two of them can sit together in addition to the fact that the three cannot sit together. In this case the answer would be 144. Can you please clarify? Agree. The question can be interpreted that way too. You can practice better questions from the link in my previous post.
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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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21 Nov 2019, 00:34
How I would solve it
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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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21 Nov 2019, 00:50
johnd32 wrote: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to each other in 6 adjacent seats in the second row of the theater. If Rob, Ray, and Susan will not sit next to each other, in how many different arrangements can the six people sit?
A. 144 B. 576 C. 714 D. 686 E. 696 Not sit together cases = Total cases  sit together cases = 6!  ( 4! x 3!) = 720  144 = 576 B is correct



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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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21 Nov 2019, 06:47
Hi everyone, thanks for the quick responses! You guys are pretty helpful. One person commented that the answer would be 144 if out of the 3 people that did not want to sit with each other, you couldn't have pairs also (meaning 2 of them sitting together and the third far away is not allowed). Can someone show me how you get to that answer 144? The question was meant to mean that neither any form of pairs nor any combination of triplets are allowed. Out of all the topics, this is my weakest. I have the 6th Ed of MGMAT prep books. IMHO, the books do a great job on all the topics EXCEPT for probability and combinatorics (too few examples and practice problems). I took the GMAT once about 3 years ago (first attempt) and got a 690 (Q: 47, V: 38, IR: 8, AWA: 5.5).I wasn't able to finish the last 4 questions on the math portion, so I randomly picked those answers. Based on the ESR, I scored between the 60th and 70th percentile for all the quant subsections, but my average time for all the subsections was above 2 minutes, so I think time management and just more practice on all the sections are needed. For my verbal, I got (percentile) 48th for CR, 93rd for RC, and 96th for SC. That's A BIG difference between my CR and other 2 subsections. I needs LOTS of help with CR. Any suggestions? I'm taking it again in 2 months and am aiming for a 780 (tough goal, I know, but I have been studying a bit). I've got only 1 shot at this, unfortunately. I don't have a job until the end of the year, so I have a bit of time to study. I've completed reviewing all the MGMAT books twice and gone over the 2018 OG questions at least twice (I've tried to take the time to not just learn the answers but how to think through them and similar questions). Can someone help with a few suggestions on drastically improving my CR score and overall quant score? Any help is appreciated. I'm fairly new to posting on this website, so if I'm not doing something right, please let me know. I know a lot of people try to help and give advice, but I feel like some of the suggestions sound good but don't really work (given by people who mean well but may not have actually done as well on the GMAT themselves, or have insufficient experience to talk about how to study well). I know different methods work for different people, but after reading so many articles and posts on many websites, I can't help but think some of the advice is misguided. Trying to sort through all of that is tough So, I'm turning to you guys for help now. Again, any help is appreciated. Thank you!



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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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21 Nov 2019, 07:59
johnd32 wrote: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to each other in 6 adjacent seats in the second row of the theater. If Rob, Ray, and Susan will not sit next to each other, in how many different arrangements can the six people sit?
A. 144 B. 576 C. 714 D. 686 E. 696 total arrangement = 6! and conditional that all three Rob, Ray, and Susan sit together = 3! and this can be arranged in 4! ways ; so we have 3!*4! = 144 not together ; 6!144 ; 576 IMO B



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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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21 Nov 2019, 08:42
johnd32 wrote: Hi everyone, thanks for the quick responses! You guys are pretty helpful. One person commented that the answer would be 144 if out of the 3 people that did not want to sit with each other, you couldn't have pairs also (meaning 2 of them sitting together and the third far away is not allowed). Can someone show me how you get to that answer 144? The question was meant to mean that neither any form of pairs nor any combination of triplets are allowed. Out of all the topics, this is my weakest. I have the 6th Ed of MGMAT prep books. IMHO, the books do a great job on all the topics EXCEPT for probability and combinatorics (too few examples and practice problems). I took the GMAT once about 3 years ago (first attempt) and got a 690 (Q: 47, V: 38, IR: 8, AWA: 5.5).I wasn't able to finish the last 4 questions on the math portion, so I randomly picked those answers. Based on the ESR, I scored between the 60th and 70th percentile for all the quant subsections, but my average time for all the subsections was above 2 minutes, so I think time management and just more practice on all the sections are needed. For my verbal, I got (percentile) 48th for CR, 93rd for RC, and 96th for SC. That's A BIG difference between my CR and other 2 subsections. I needs LOTS of help with CR. Any suggestions? I'm taking it again in 2 months and am aiming for a 780 (tough goal, I know, but I have been studying a bit). I've got only 1 shot at this, unfortunately. I don't have a job until the end of the year, so I have a bit of time to study. I've completed reviewing all the MGMAT books twice and gone over the 2018 OG questions at least twice (I've tried to take the time to not just learn the answers but how to think through them and similar questions). Can someone help with a few suggestions on drastically improving my CR score and overall quant score? Any help is appreciated. I'm fairly new to posting on this website, so if I'm not doing something right, please let me know. I know a lot of people try to help and give advice, but I feel like some of the suggestions sound good but don't really work (given by people who mean well but may not have actually done as well on the GMAT themselves, or have insufficient experience to talk about how to study well). I know different methods work for different people, but after reading so many articles and posts on many websites, I can't help but think some of the advice is misguided. Trying to sort through all of that is tough So, I'm turning to you guys for help now. Again, any help is appreciated. Thank you! Hi, Johnd32, Go through the below link. It may help you a lot. https://gmatclub.com/forum/expertstopi ... 43170.html (Ninja's CR guide) Regards, Raxit T.
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Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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21 Nov 2019, 10:10
This is the totally wrong thread to ask a question like this but in general I would advise you to lower your expectancies, have you ever come close such a score in your Mocks? Scoring a 780 puts you in the leauge of veteran GMAT tutors who been teaching, living and breathing this sh*t full time  given that your previous take wasn't even close to that skill level I hardly doubt that you should even think about a 780. I dont even put you down, but I dont see why anyone, besides people who probably scored 700 on their diagnostic with ease, should put such a high burden on themselves and straight aim for a 780. Scoring anywhere from 700750 would open you all doors and you would not have to be disappointed at all if you actually manage to score somewhere in that range. It seems that you somewhat have a distorted expectancy of what is actually achievable.
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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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21 Nov 2019, 10:38
BunuelCan someone please help me understand why the method below is wrong. The general consensus seems to be that the answer is 576, not 144. What am I doing wrong below? If anyone else can also help point it out (if Bunuel is busy), it'd be much appreciated. Thanks everyone. Option 1
If 3 of them are to not sit next to each other, those three can sit in the following set up. In the diagram below, Y denotes a seat filled with someone who does not want to sit next to another person, and _ denotes an empty seat which can be filled by those without seating restrictions.
Y _ Y _ Y _ Y _ _ Y _ Y Y _ Y _ _ Y _ Y _ Y _ Y
FOR EACH of the options above, the 3 of them who need to be separated can sit in 3! ways = 6 ways. In addition to that, FOR EACH of the options, the remaining empty seats can also be filled in 3! ways = 6 ways.
Hence, the ways the 3 of them won't sit next to each other is 4*6*6 = 144



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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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21 Nov 2019, 10:43
BunuelActually, if 2 of them cannot sit next to each other, the number of arrangements is 480. If three of them cannot sit next to each other (more constraints), the answer should be smaller, not 576 (larger), right?



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Re: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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24 Nov 2019, 18:39
johnd32 wrote: Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to each other in 6 adjacent seats in the second row of the theater. If Rob, Ray, and Susan will not sit next to each other, in how many different arrangements can the six people sit?
A. 144 B. 576 C. 714 D. 686 E. 696 The total number of ways to seat all six people is 6! = 720. When Rob, Ray, and Susan sit next to each other, we have: [RYS][L][M][J] We see that we have 4 total spots, or 4!, and we must account for the number of ways to arrange RYS, which is 3!. So, the total number of ways to arrange the group when Rob, Ray, and Susan sit next to each other is 3! x 4! = 6 x 24 = 144. So, the total number of ways to arrange the group when Rob, Ray, and Susan do not sit next to each other is 720  144 = 576, Answer: B
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Rob, Lewis, Ray, Susan, Mary, and Janet go to a movie and sit next to
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27 Nov 2019, 11:15
johnd32 wrote: There are 2 ways that come to mind to solve this, but I'm getting different answers. I've been trying to figure this out for a while. Just can't see where I'm going wrong.
Option 1
If 3 of them are to not sit next to each other, those three can sit in the following set up. In the diagram below, Y denotes a seat filled with someone who does not want to sit next to another person, and _ denotes an empty seat which can be filled by those without seating restrictions.
Y _ Y _ Y _ Y _ _ Y _ Y Y _ Y _ _ Y _ Y _ Y _ Y
FOR EACH of the options above, the 3 of them can sit in 3! ways = 6 ways. In addition to that, FOR EACH of the options, the remaining empty seats can also be filled in 3! ways = 6 ways.
Hence, the ways the 3 of them won't sit next to each other is 4*6*6 = 144
Option 2
Calculate total number of seating arrangements  Ways 3 of them can sit together
Total number of seating arrangements = 6! = 720
For ways 3 of them can sit together, pretend the group is a group of 4 individuals instead of 6 (because 3 of them have to sit together anyway). If you had 4 individuals, you'd have 4! ways = 24 ways. However, one of those individuals is actually a group of 3 people, and the ways that group of 3 people can sit is 3! = 6. This 6 multiplied by the 24 calculated earlier gives you the number of ways 3 of them can sit together, which is 144.
Total seating arrangements  Ways of sitting together = 720  144 = 576
I'm getting different answers in Option 1 and Option 2, and I'm not sure if either of them is even correct. Any help would be appreciated. Thanks everyone! The reason you are getting different answers for Option 1 and Option 2 is because each of those solutions are answering a different question. The question is not very clear on what is meant by "Rob, Ray, and Susan will not sit next to each other". If this means no two of them can sit together (for instance, RRLMJS is not acceptable), then the number of different arrangements is 144. Your first option is correctly answering this question. If "Rob, Ray, and Susan will not sit next to each other" means that two of them can sit together as long as all three are not sitting together (i.e. RRLMJS is acceptable), then the number of different arrangements is 576. Your second option is answering this question. If you wanted to get the answer of 576 using the method of your first option, you would also have to take into account the arrangements such as YY_Y_ _, Y_YY_ _ etc. If you wanted to get the answer of 144 using the method of your second option, you need to calculate the number of arrangements where at least two of the three people are sitting together. If we consider Rob and Ray sitting together, for instance, the number of arrangements is 2*5!. The same is true when Rob and Susan sit together, as well as when Ray and Susan sit together. When you add all these arrangements, you will have double counted the arrangements when all three of them sit together and we know the number of such arrangements is 3!*4!. Thus, in 2*5! + 2*5! + 2*5!  3!*4! = 6*5!  6*4! = 6*4!(5  1) = 576 of the arrangements, at least two of the three people are sitting together and thus, in 720  576 = 144 of the arrangements, no two of those people are sitting together.
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