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There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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08 Aug 2011, 13:00
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There are 6 people at a party sitting at a round table with 6 seats: A, B, C, D, E and F. A CANNOT sit next to either D or F. How many ways can the 6 people be seated? A. 720 B. 120 C. 108 D. 84 E. 48
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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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08 Aug 2011, 14:38
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When items are arranged in circle the permutation formula is (n1)!
so in this case, let us first calculate total number of ways in which 6 people can be arranged. from permutation formula when items are arranged in circle we have (61)! possible permutation. (61)! = 120.
But we have restrictions that A cannot seat next to d or F. let us calculate the ways in which A can seat next to D. If AD sits together then 4 other people have to be arranged. so again we have apply permutation formula for 4 items in circular arrangements. we get (41)! = 3! =6. so A and D sit next to each other in 6 ways, similarly A and F sit next to each other in 6 ways. so to arrive at our answer we have to subtract ways in which A sits next to D & F from total number of ways these six people are sitting. so our answer is 120 (6 +6) = 108.
edited I made an error in calculating in this post and this solution is wrong.
Last edited by Aj85 on 08 Aug 2011, 16:08, edited 1 time in total.



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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08 Aug 2011, 15:22
Aj85 wrote: When items are arranged in circle the permutation formula is (n1)!
so in this case, let us first calculate total number of ways in which 6 people can be arranged. from permutation formula when items are arranged in circle we have (61)! possible permutation. (61)! = 120.
But we have restrictions that A cannot seat next to d or F. let us calculate the ways in which A can seat next to D. If AD sits together then 4 other people have to be arranged. so again we have apply permutation formula for 4 items in circular arrangements. we get (41)! = 3! =6. so A and D sit next to each other in 6 ways, similarly A and F sit next to each other in 6 ways. so to arrive at our answer we have to subtract ways in which A sits next to D & F from total number of ways these six people are sitting. so our answer is 120 (6 +6) = 108.
Answer C. A B C D E F can be seated in 5! ways. When you consider AD as 1, total number of people becomes AD  B  C  E  F > 5 not 4 Similarly, when AF is 1, total number of people AF B C D E > 5 not 4 There would also be a case where DAF will sit together, and we have to eliminate that case from the total number of combinations because we have included it twice when A sits with D and F each time. Please help explain this, your answer may be correct but " If AD sits together then 4 other people have to be arranged" AD as a person has to be arranged too.



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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08 Aug 2011, 16:04
sap wrote: A B C D E F can be seated in 5! ways.
When you consider AD as 1, total number of people becomes AD  B  C  E  F > 5 not 4 Similarly, when AF is 1, total number of people AF B C D E > 5 not 4
There would also be a case where DAF will sit together, and we have to eliminate that case from the total number of combinations because we have included it twice when A sits with D and F each time.
Please help explain this, your answer may be correct but "If AD sits together then 4 other people have to be arranged" AD as a person has to be arranged too. Thanks Sap for pointing out my error, I did rechecked the question . Actually I learned that when it is circular arrangement we subtract 1, so I carelessly subtracted 1 from each and every stage where i had to compute permutation including not counting ad as one item. So the solution will be adcdef = (61)! = 120. Then ad(another 4people) = (51)! = 24 similarly AF(another 4 people) = (51)! =24. so we have to deduct 24 +24 from 120 which equals to 72. But we have double counted daf options. so we have to add cases where DAF are toghter. DAF (another 3peopl) = (41)! = 6. DAF itself will have 2 permutations, so we multiply 6 by 2 which equals to 12. Now we add 72+ 12 =84. so now the answer by this method = 84. I hope some experts solve this question correctly.



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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08 Aug 2011, 21:57
Aj85 wrote: sap wrote: A B C D E F can be seated in 5! ways.
When you consider AD as 1, total number of people becomes AD  B  C  E  F > 5 not 4 Similarly, when AF is 1, total number of people AF B C D E > 5 not 4
There would also be a case where DAF will sit together, and we have to eliminate that case from the total number of combinations because we have included it twice when A sits with D and F each time.
Please help explain this, your answer may be correct but "If AD sits together then 4 other people have to be arranged" AD as a person has to be arranged too. Thanks Sap for pointing out my error, I did rechecked the question . Actually I learned that when it is circular arrangement we subtract 1, so I carelessly subtracted 1 from each and every stage where i had to compute permutation including not counting ad as one item. So the solution will be adcdef = (61)! = 120. Then ad(another 4people) = (51)! = 24 similarly AF(another 4 people) = (51)! =24. so we have to deduct 24 +24 from 120 which equals to 72. But we have double counted daf options. so we have to add cases where DAF are toghter. DAF (another 3peopl) = (41)! = 6. DAF itself will have 2 permutations, so we multiply 6 by 2 which equals to 12. Now we add 72+ 12 =84. so now the answer by this method = 84. I hope some experts solve this question correctly. I solved it in the same way and got 84, but the answer is C. If the question is: A cannot sit next to both D and F(A cannot sit between D and F), then the answer would be 108. Also, I saw another explanation on the forum fix a position for A. It has two adjacent positions one on either side. D and F cannot take these two positions. So, D and F can be placed in the remaining 3 positions in 3P2 = 6ways. The remaining three positions can be filled by B,C,E in 3! ways. So, total number of ways = 6*6 = 36 ways. I don't know how good is this explanation in a circular arrangement... Do share your thoughts.



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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08 Aug 2011, 22:09
Aj85 wrote: sap wrote: A B C D E F can be seated in 5! ways.
When you consider AD as 1, total number of people becomes AD  B  C  E  F > 5 not 4 Similarly, when AF is 1, total number of people AF B C D E > 5 not 4
There would also be a case where DAF will sit together, and we have to eliminate that case from the total number of combinations because we have included it twice when A sits with D and F each time.
Please help explain this, your answer may be correct but "If AD sits together then 4 other people have to be arranged" AD as a person has to be arranged too. Thanks Sap for pointing out my error, I did rechecked the question . Actually I learned that when it is circular arrangement we subtract 1, so I carelessly subtracted 1 from each and every stage where i had to compute permutation including not counting ad as one item. So the solution will be adcdef = (61)! = 120. Then ad(another 4people) = (51)! = 24 similarly AF(another 4 people) = (51)! =24. so we have to deduct 24 +24 from 120 which equals to 72. But we have double counted daf options. so we have to add cases where DAF are toghter. DAF (another 3peopl) = (41)! = 6. DAF itself will have 2 permutations, so we multiply 6 by 2 which equals to 12. Now we add 72+ 12 =84. so now the answer by this method = 84. I hope some experts solve this question correctly. Guys, in addition to above explanation You have subtracted the cases AD and AF but you have neglected the cases DA and FA. Each of which will be having 24 cases. So, you need to subtract 2*24 from 84 which will give 8448 = 36.



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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08 Aug 2011, 22:32
Viv, I didnt count ad and da because i thought they were in circles, I mean in linear arrangement if we have A B and C we can arrange like ABC, ACB, BAC, ACA, CAB and CBA, but when they are in circle ABC and BCA and CAB are same, That's what I learnt for circular arrangements. of course I may be wrong. But its very interesting question, hope someone solves it correctly . It will be great to see a correct solution for this question.



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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08 Aug 2011, 23:43
Did you get this from the Manhattan book, or from one of their MGMAT s ?



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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10 Aug 2011, 07:57
Aj85 wrote: Viv, I didnt count ad and da because i thought they were in circles, I mean in linear arrangement if we have A B and C we can arrange like ABC, ACB, BAC, ACA, CAB and CBA, but when they are in circle ABC and BCA and CAB are same, That's what I learnt for circular arrangements. of course I may be wrong. But its very interesting question, hope someone solves it correctly . It will be great to see a correct solution for this question. Aj85, You are correct to say that, in circular arrangements ABC, ACB, BAC, ACA, CAB and CBA will represent same arrangement. Hence one we have subtracted AF and AD we need not subtract FA and DA. I am too getting the answer as 84. @sap can U please let us know the source and the OA



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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12 Aug 2011, 20:21
Sudhanshuacharya wrote: Aj85 wrote: Viv, I didnt count ad and da because i thought they were in circles, I mean in linear arrangement if we have A B and C we can arrange like ABC, ACB, BAC, ACA, CAB and CBA, but when they are in circle ABC and BCA and CAB are same, That's what I learnt for circular arrangements. of course I may be wrong. But its very interesting question, hope someone solves it correctly . It will be great to see a correct solution for this question. Aj85, You are correct to say that, in circular arrangements ABC, ACB, BAC, ACA, CAB and CBA will represent same arrangement. Hence one we have subtracted AF and AD we need not subtract FA and DA. I am too getting the answer as 84. @sap can U please let us know the source and the OA ok, here's an attempt A,B,C,D,E and F can sit in (61)! ways  120 Now consider this, when we take AD as one person all of them can sit in AD, B, C, D ,E > (51)! > 24 now, this calculation considers clockwise and anti clockwise as different, i.e AD is different from DA. Therefore we have to divide this figure by 2, because We do not want to distinguish between AD and DA. Hence there are 12 cases when AD sit to each other ( irrespective of AD or DA ) Similarly, there are 12 cases when AF sit to each other. Similarly, we can say ADF can also be clockwise and anticlockwise, but we have to realize that, when counting AD and AF we have counted DAF or FAD TWICE ( once for each ) therefore we need to deduct the arrangement for DAF only once. If DAF were 1 person, possible number of arrangements  ADF,B,C,E > 3! We will not count it twice because we need to deduct this only once from AD and AF combination ! Therefore total count 5!  (4!/2 + 4!/2 3!) = 120  (12 + 12 12 ) = 108 what say ?



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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05 Oct 2011, 21:12
i just came across this problem in the manhattan review turbocharge math. answer key says C. 108. The solutions manual has an explanation that I don't really understand.



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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05 Oct 2011, 22:08
mrdanielkim wrote: i just came across this problem in the manhattan review turbocharge math. answer key says C. 108. The solutions manual has an explanation that I don't really understand. can you please put the explanation here ?



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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05 Oct 2011, 22:47
i figured it out after taking a break from studying. thanks anyway!
here's the solution anyway, though the book has the question worded differently:
Q: IN how many ways can 6 people be seated at a round table if one of those seated cannot sit next to two of the other five:
A: Six people can be seated around a round table in 5! ways. there are 2 ways that the two unwelcome people could sit next to the person in question and 3! ways of arranging the other three. This is subtracted from the base case of 5!, giving the result of 108.



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Re: There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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06 Oct 2011, 04:01
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mrdanielkim wrote: i figured it out after taking a break from studying. thanks anyway!
here's the solution anyway, though the book has the question worded differently:
Q: IN how many ways can 6 people be seated at a round table if one of those seated cannot sit next to two of the other five:
A: Six people can be seated around a round table in 5! ways. there are 2 ways that the two unwelcome people could sit next to the person in question and 3! ways of arranging the other three. This is subtracted from the base case of 5!, giving the result of 108. This question is different from the one posted above. There are two versions and the answer would be different in the two cases. Let me pick this version first: There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, cannot sit next to D and F at the same time. (This means that A can sit next to D but not while F is on A's other side. Similarly, A can sit next to F too but not while D is on A's other side) Total number of ways of arranging 6 people in a circle = 5! = 120 In how many of these 120 ways will A be between D and F? We make DAF sit on three consecutive seats and make other 3 people sit in 3! ways. or we make FAD sit of three consecutive seats and make other 3 people sit in 3! ways. In all, we make A sit next to D and F simultaneously in 12 ways. 120  12 = 108 is the number of ways in which D and F are not sitting next to A at the same time. The second version which seemed like the intended meaning of the original poster: There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, can sit neither next to D nor next to F. (This means that A cannot sit next to D in any case and A cannot sit next to F in any case.) Here, we say that A has to sit next to two of B, C and E. Let's choose 2 of B, C and E in 3C2 = 3 ways. Let's arrange them around A in 2 ways (say we choose B and C. We could have BAC or CAB). We make these 3 sit on any 3 consecutive seats in 1 way. Number of ways of arranging these 3 people = 3*2 = 6 The rest of the 3 people can sit in 3! = 6 ways Total number of ways in which A will sit neither next to D nor next to F = 6*6 = 36 ways
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There are 6 people at a party sitting at a round table with 6 seats: A [#permalink]
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17 Oct 2016, 08:10
vivgmat wrote: There are 6 people at a party sitting at a round table with 6 seats: A, B, C, D, E and F. A CANNOT sit next to either D or F. How many ways can the 6 people be seated?
A. 720 B. 120 C. 108 D. 84 E. 48 Let’s fix D in the first position, and F in the second. D F _ _ _ _ A cannot be in positions 3 nor 6, thus in position 3 we have three options: B, C, or E. In position 6 we have the other two choices. In position 4 we have two options: the remaining choice and A. And for position 5 we have only one. D F 3 2 1 2 => 12 Let’s fix D in the first position, and F in the third. D _ F _ _ _ A can only be on position 5 => D _ F _ A _ , So we have D 3 F 2 A 1 => 6 Let’s fix D in the first position, and F in the fourth, D _ _ F_ _ = > not possible Let’s fix D in the first position, and F in the fifth. D _ _ _ F _ => D 3 A 2 F 1 => 6 Let’s fix D in the first position, and F in the sixth. D _ _ _ _ F A cannot be in positions 2 nor 5 => D 3 2 1 2 F => 12 So, 12 + 6 + 6 + 12 = 36 Correct answer: 36 . None of the aboveAnother way: Let’s fix A in position 2 => _ A _ _ _ _ So neither D nor F can be in either position 1 or 3. So in position 1 we have 3 choices: B, C, or E. In position 3 we have the other two choices. In position 4 we have three options: the remaining choice, D, and F. In position 5 two options and in 6 one. 3 A 2 3 2 1 => 36 Correct answer: 36 . None of the above
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