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Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and
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Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements
Originally posted by GMAT100 on 15 Feb 2007, 18:09.
Last edited by GMAT100 on 15 Feb 2007, 21:11, edited 1 time in total.




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Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and
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15 Dec 2013, 04:47
GMAT100 wrote: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements The best way to deal with the questions like this is to find the total # of arrangements and then subtract # of arrangements for which opposite of restriction occur: Total # of arrangements of 6 people (let's say A, B, C, D, E, F) is \(6!\). # of arrangement for which 2 particular persons (let's say A and B) are adjacent can be calculated as follows: consider these two persons as one unit like {AB}. We would have total 5 units: {AB}{C}{D}{E}{F}  # of arrangement of them 5!, # of arrangements of A and B within their unit is 2!, hence total # of arrangement when A and B are adjacent is \(5!*2!\). # of arrangement when A and B are not adjacent is \(6!5!*2!=480\).
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total ways of arranging 6 ppl = 6! = 6*5*4*3*2 = 720
ways in which m&j sit together = 5!*2 = 240
ans = 720  240 = 480




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480 as well
total number of arrangements 6*5*4*3*2*1 = 720
want to count criteria we dont want
If Marcia sits in first chair, and Jan sits in 2nd chair there are 4! = 24 arrangements for the other people
Jan could also sit in first chair and Marcia in 2nd, so there are also 24 different arrangments
24+24 = 48
so we can have 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or 5th and 6th
therefore there are 5 ways each have 48 combinations so 48*5=240
720240 = 480



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Re: Advanced Constraint Combinatorics
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04 Aug 2011, 21:12
GMAT100 wrote: The following Question:
Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theatre. If Marcia and Jan will not sit next to each other, in how many different arrangements
Answer will be posted tomorrow @krishp84 krishp84 wrote: why cannot we apply symmetry here ?
We cannot apply symmetry because the number of cases where M and J are sitting together is not equal to the number of cases where M and J are not sitting together. Consider 3 people: A, B and C They can be arranged in 3! = 6 ways ABC ACB BAC BCA CAB CBA In how many of these are A and B sitting together? 2 In how many are they not sitting together? 4 When we arrange 6 people here in 6! (= 720) ways, in how many of those will M and J sit together? Consider M and J to be one and arrange 5 people in 5! ways. Also, M and J can exchange places so multiply by 2. You get 5!*2 = 240 ways Out of 720, M and J will be next to each other in only 240 ways. krishp84 wrote: I recall a question related to this  "6 people standing in a line and one person(A) cannot stand behind another person(B). What is the total number of ways of forming the line? " Cannot remember the complete question, something like this...But remember that it was pure symmetry : 6!/2=360 ways
That question would be something like this: 6 people go to a movie and sit next to each other in 6 adjacent seats in the front row of the theatre. If Marcia will not sit to the right of Jan, how many different arrangements are possible? 'to the right of Jan' means anywhere on the right, not necessarily on the adjacent seat. Here we see symmetry because there are only 2 ways in which Marcia can sit. She can sit either to the left of Jan (any seat on the left) or to the right of Jan (any seat on the right). There is nothing else possible. The number of cases in which she will sit to the left will be same as the number of cases in which she will sit to the right. That is why the answer here will be 6!/2. But the original question talks about sitting right next to each other on adjacent seats. The probability of sitting 'next to each other' is less than the probability of sitting 'not next to each other'. Hence we cannot apply the symmetry principle there.
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Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and
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05 Aug 2011, 17:39
Variant2.1 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ?
Variant3.1 of this question 
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different arrangements ?
Variant4.1 of this question 
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ?



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Re: Advanced Constraint Combinatorics
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16 Aug 2011, 22:28
Variant1 of this question  7 people(A,B,C,D,E,F.H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different arrangements ? > 7!  2((72+1)!) = 7!2(6!)=7(6!)2(6!) = 5(6!) ways Yes, that's fine. You make them sit together and subtract that out of 7!.Variant2 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A,F and E will not sit next to each other in how many different arrangements ? 7!(3!)(73+1)! = 7! 6(5!) ways A little ambiguity here. I would say  A, F and E will not all sit together in how many ways? The solution is fine.
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Re: Advanced Constraint Combinatorics
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16 Aug 2011, 22:32
Variant3 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different arrangements ? > 8P72[(82+1)P(72+1)] = 8P72(7P6) ways = 8!  2(7!) = 6(7!) ways The answer is correct. An alternative approach could be this:
I will just think of the vacant seat as a person V. So 8 people, 8 seats in 8! ways. 2 people should not be together so 8!  2*7!
Another interesting variant could be this: 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one person between A and F?
Try it.Variant4 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A ,F and E will not sit next to each other in how many different arrangements ? Correct solution. Alternative approach parallel to the one above:
The vacant spot is a person V. Required arrangements: 8!  3!*6!
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Re: Advanced Constraint Combinatorics
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16 Aug 2011, 22:42
Variant1.1 of this question  7 people(A,B,C,D,E,F.H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different arrangements ? > 7!/2 ways Correct.
Variant2.1 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ? Variant3.1 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different arrangements ? Variant4.1 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ? You can apply symmetry in all these questions. Tell me how you will do it. I will confirm the answer.
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Re: Advanced Constraint Combinatorics
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18 Aug 2011, 18:30
VeritasPrepKarishma wrote: Variant3 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different arrangements ?
>
8P72[(82+1)P(72+1)] = 8P72(7P6) ways = 8!  2(7!) = 6(7!) ways
The answer is correct. An alternative approach could be this:
I will just think of the vacant seat as a person V. So 8 people, 8 seats in 8! ways. 2 people should not be together so 8!  2*7! WoW  Loved this Approach > Kudos +1VeritasPrepKarishma wrote: Another interesting variant could be this: 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one person between A and F?
Try it. This is equivalent to saying 8 people sit in 8 adjacent seats with an imaginary girl Ms. X with them Subtract the number of arrangements when Ms. X sits between A and F AXF FXA XAF XFA AFX FAX All the above are equivalent  No person sits between A and F So answer will be 8!(3!)(6!) waysNow let me apply my traditional approach and confirm this  8P7(3!)((83+1)P(73+1)) = 8P7(3!)(6P5) = 8!(3!)(6!) ways



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Re: Advanced Constraint Combinatorics
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18 Aug 2011, 18:43
krishp84 wrote: Variant2.1 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theater. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ? Let me post my solution  This will be equivalent as below 
A can come to left/right of F F can come to left/right of E (I diagrammed, but it is timetaking to create a picture and post here)
So number of different arrangements = (1/2)(7!/2) = 7!/4 ways



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Re: Advanced Constraint Combinatorics
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18 Aug 2011, 18:56
VeritasPrepKarishma wrote: Variant3.1 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different arrangements ? This will be 8!/2 ways or (8P7)/2 = 8!/2 waysVeritasPrepKarishma wrote: Variant4.1 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the left of E in how many different arrangements ? This will be (1/2)(8!/2) = 8!/4 ways or (1/2)(1/2)(8P7) = 8!/4 waysLogic  combination of variant 2.1 and 3.1 VeritasPrepKarishma wrote: You can apply symmetry in all these questions. Tell me how you will do it. I will confirm the answer. I understand this is a HUGE post  But could not find another way to link all these related concepts.Welcome to any new variant. I LOVE these puzzles.



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Re: Advanced Constraint Combinatorics
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18 Aug 2011, 22:39
VeritasPrepKarishma wrote: Variant3 of this question  7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different arrangements ?
>
I will just think of the vacant seat as a person V. So 8 people, 8 seats in 8! ways. 2 people should not be together so 8!  2*7!
Another interesting variant could be this: 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one person between A and F?
krishp84 wrote: This is equivalent to saying 8 people sit in 8 adjacent seats with an imaginary girl Ms. X with them Subtract the number of arrangements when Ms. X sits between A and F AXF FXA XAF XFA AFX FAX All the above are equivalent  No person sits between A and F So answer will be 8!(3!)(6!) ways This variant wants you to put at least one person between A and F. This means that all those cases where A and F are together are not acceptable and also those cases where A and F have V (or your Ms. X!) between them are not acceptable. Number of cases where A and F are together = 2*7! Number of cases where A and F have V between them = 2*6! [AVF and FVA] These are the cases that are not acceptable. Answer should be 8!  2*7!  2*6! = 8!  16*6! Think about it another way: Compare the two questions. One where you don't want them to be together, the other where you don't want them to be together and you don't want the vacant spot between them. Obviously, in the second case, the number of cases you do not want are higher. In the first question, you subtracted a total of 2*7! arrangements i.e. 14*6! arrangements. In the second question, you need to subtract some more. You cannot subtract 3!*6! = 6*6! arrangements only.
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Re: Advanced Constraint Combinatorics
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18 Aug 2011, 22:47
krishp84 wrote: I understand this is a HUGE post  But could not find another way to link all these related concepts.[/color] Welcome to any new variant. I LOVE these puzzles. Your solutions are correct. In a few weeks, I am going to start Permutation Combination on my blog. I already intend to make a post on this question and all its variants. If you want to try out further complications, try putting in 2 or 3 vacant spots. Now there will be 23 identical people named 'V' so adjustments will be needed.
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Re: Advanced Constraint Combinatorics
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19 Aug 2011, 16:08
VeritasPrepKarishma wrote: This variant wants you to put at least one person between A and F. This means that all those cases where A and F are together are not acceptable and also those cases where A and F have V (or your Ms. X!) between them are not acceptable. Number of cases where A and F are together = 2*7! Number of cases where A and F have V between them = 2*6! [AVF and FVA] These are the cases that are not acceptable. Answer should be 8!  2*7!  2*6! = 8!  16*6!
Think about it another way: Compare the two questions. One where you don't want them to be together, the other where you don't want them to be together and you don't want the vacant spot between them. Obviously, in the second case, the number of cases you do not want are higher. In the first question, you subtracted a total of 2*7! arrangements i.e. 14*6! arrangements. In the second question, you need to subtract some more. You cannot subtract 3!*6! = 6*6! arrangements only. Yes  I made that mistake of assuming X/V as a part of the group with A,F instead of thinking them separately. Stupid me !!!



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Re: Advanced Constraint Combinatorics
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19 Aug 2011, 16:12
VeritasPrepKarishma wrote: If you want to try out further complications, try putting in 2 or 3 vacant spots. Now there will be 23 identical people named 'V' so adjustments will be needed. Yes  I was thinking on these lines....I will post some variants on this once I am free.Also thought of combining Probability with this because it is so related especially  the symmetry part.



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Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and
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17 Aug 2014, 15:39
Bunuel wrote: GMAT100 wrote: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements The best way to deal with the questions like this is to find the total # of arrangements and then subtract # of arrangements for which opposite of restriction occur: Total # of arrangements of 6 people (let's say A, B, C, D, E, F) is \(6!\). # of arrangement for which 2 particular persons (let's say A and B) are adjacent can be calculated as follows: consider these two persons as one unit like {AB}. We would have total 5 units: {AB}{C}{D}{E}{F}  # of arrangement of them 5!, # of arrangements of A and B within their unit is 2!, hence total # of arrangement when A and B are adjacent is \(5!*2!\). # of arrangement when A and B are not adjacent is \(6!5!*2!=480\). Hi , i want to refer to another question that you answered. it is on this link : alicialivesinatownwhosestreetsareonagridsystem126920.htmlmy question is : on this question above. when you glued AB , the total units changed from 6 units to 5 units {AB}{C}{D}{E}{F}. but on the other question , you glued the (SS) and the units changed from 6 units to 4 (SEEE). this doesn't make sense to me. can you explain that please ….. thanks



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Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and
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18 Aug 2014, 04:34
shagalo wrote: Bunuel wrote: GMAT100 wrote: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements The best way to deal with the questions like this is to find the total # of arrangements and then subtract # of arrangements for which opposite of restriction occur: Total # of arrangements of 6 people (let's say A, B, C, D, E, F) is \(6!\). # of arrangement for which 2 particular persons (let's say A and B) are adjacent can be calculated as follows: consider these two persons as one unit like {AB}. We would have total 5 units: {AB}{C}{D}{E}{F}  # of arrangement of them 5!, # of arrangements of A and B within their unit is 2!, hence total # of arrangement when A and B are adjacent is \(5!*2!\). # of arrangement when A and B are not adjacent is \(6!5!*2!=480\). Hi , i want to refer to another question that you answered. it is on this link : alicialivesinatownwhosestreetsareonagridsystem126920.htmlmy question is : on this question above. when you glued AB , the total units changed from 6 units to 5 units {AB}{C}{D}{E}{F}. but on the other question , you glued the (SS) and the units changed from 6 units to 4 (SEEE). this doesn't make sense to me. can you explain that please ….. thanks In other question we are interested in routs which start with {SS}. We are fixing {SS} and counting the number of permutations of the remaining letters SEEE. Please reread the solution.
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Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and
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25 Nov 2016, 05:23
Total number of arrangements =6!= 720 Total arrangements of J and M sitting together = 5!*2 = 240 Total arrangement of J and M not sitting together = 480 Answer = 480
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Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and
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02 Feb 2018, 12:06
GMAT100 wrote: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements We can use the following formula: Total number of ways to arrange the 6 people = (number of arrangements whenMarcia sits next to Jan) + (number of arrangements when Marica does not sit next to Jan) Let’s determine the number of arrangements in which Marcia sits next to Jan. We can denote Greg, Marcia, Peter, Jan, Bobby and Cindy as G,M, P, J, B, and C, respectively. If Marcia and Jan must sit together, we can consider them as one person [MJ]. For example, one seating arrangement could be [MJ][G][P][B][C]. Thus, the number of ways to arrange five people in a row is 5! = 120. However, we must also account for the ways we can arrange Marcia and Jan, that is, either [MJ] or [JM]. Thus, there are 2! = 2 ways to arrange Marcia and Jan. Therefore, the total number of seating arrangements is 120 x 2 = 240 if Marcia and Jan DO sit next to each other. Since there are 6 people being arranged, the total number of possible arrangements is 6! = 720. Thus, the number of arrangements when Marcia does NOT sit next to Jan is 720  240 = 480.
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