December 11, 2018 December 11, 2018 09:00 PM EST 10:00 PM EST Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST. December 13, 2018 December 13, 2018 08:00 AM PST 09:00 AM PST What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
Author 
Message 
TAGS:

Hide Tags

Manager
Status: Only GMAT!!
Joined: 17 Sep 2010
Posts: 64
WE 1: 5.5+ years IT Prof.

There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
08 Aug 2011, 12:00
Question Stats:
40% (02:14) correct 60% (02:35) wrong based on 240 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8659
Location: Pune, India

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
06 Oct 2011, 03:01
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
_________________
[b]Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >




Manager
Joined: 14 Mar 2011
Posts: 192

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
Updated on: 08 Aug 2011, 15:08
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.
Originally posted by Aj85 on 08 Aug 2011, 13:38.
Last edited by Aj85 on 08 Aug 2011, 15:08, edited 1 time in total.



Intern
Joined: 22 Sep 2010
Posts: 23

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
08 Aug 2011, 14:22
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.



Manager
Joined: 14 Mar 2011
Posts: 192

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
08 Aug 2011, 15:04
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.



Manager
Status: Only GMAT!!
Joined: 17 Sep 2010
Posts: 64
WE 1: 5.5+ years IT Prof.

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
08 Aug 2011, 20:57
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.



Manager
Status: Only GMAT!!
Joined: 17 Sep 2010
Posts: 64
WE 1: 5.5+ years IT Prof.

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
08 Aug 2011, 21:09
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.



Manager
Joined: 14 Mar 2011
Posts: 192

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
08 Aug 2011, 21:32
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.



Manager
Joined: 26 Oct 2010
Posts: 81

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
08 Aug 2011, 22:43
Did you get this from the Manhattan book, or from one of their MGMAT s ?



Manager
Joined: 31 May 2011
Posts: 79
Location: India
Concentration: Finance, International Business
GMAT Date: 12072011
GPA: 3.22
WE: Information Technology (Computer Software)

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
10 Aug 2011, 06:57
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



Intern
Joined: 22 Sep 2010
Posts: 23

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
12 Aug 2011, 19:21
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 ?



Intern
Joined: 15 Sep 2009
Posts: 2

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
05 Oct 2011, 20:12
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.



Intern
Joined: 22 Sep 2010
Posts: 23

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
05 Oct 2011, 21:08
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 ?



Intern
Joined: 15 Sep 2009
Posts: 2

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
05 Oct 2011, 21:47
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.



Manager
Status: GMAT Coach
Joined: 05 Nov 2012
Posts: 130
Location: Peru
GPA: 3.98

There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
17 Oct 2016, 07:10
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
_________________
Clipper Ledgard GMAT Coach



Director
Joined: 08 Jun 2013
Posts: 514
Location: India
GPA: 3.82
WE: Engineering (Other)

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
23 Aug 2018, 07:32
chetan2u, VeritasKarishma, Bunuel, mikemcgarry, niks18, generisModerators.. can you pl update the OA for this question?
_________________
It seems Kudos button not working correctly with all my posts...
Please check if it is working with this post......
is it?....
Anyways...Thanks for trying



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8659
Location: Pune, India

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
23 Aug 2018, 20:58
Harshgmat wrote: chetan2u, VeritasKarishma, Bunuel, mikemcgarry, niks18, generisModerators.. can you pl update the OA for this question? The correct answer is not there in any of the options. If it is a Manhattan question, the exact wording of the original question needs to be put.
_________________
[b]Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 16 Jun 2018
Posts: 25
GPA: 4
WE: Supply Chain Management (Energy and Utilities)

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
24 Aug 2018, 09:50
A can be seated along the circular table having six seats in 6 ways. When A has occupied one seat, we are left with only 3 seats where D & F can be seated (As adjacent seats are restricted for D&F). These 2 persons can be seated in 3 seats in 3X2 ways = 6 Ways. (D can take any seat among 3 seats left and then F can take any 2 seats left). Finally we have 3 seats left for B, C & E and number of arrangements would be 3X2X1 = 6 Ways. Total Number of ways 6 X 6 X 6 = 216 Ways. (I am unable to figure out where I am doing wrong, Please help me with this issue).



Intern
Joined: 03 Apr 2017
Posts: 45

Re: There are 6 people at a party sitting at a round table with 6 seats: A
[#permalink]
Show Tags
24 Aug 2018, 13:19
VeritasKarishma wrote: 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 I was trying to do the second (intended meaning's calculation, albeit in a different way. I was getting a different answer. Can you tell me where I am going wrong? SO let us take the total number of arrangements: (61)! = 5! = 120 Cases where A & F are together: (4!) * 2 (Considering that AF and FA are different) = 48 Cases where A & D are together: (4!) * 2 (Considering that AD and DA are different) = 48 Now the above 2 cases will each include identical cases where a D was immediately after or before AF &FA or an F was immediately after or before AD and DA. Each case will be counted twice. Number of such cases counted twice is ADF taken together and permutated i.e. 3! * 3! (first factorial for the circular permutation, and the second for the permutation among ADF) = 36 now 1204848+36 = 60 should be the answer according to me. I know this is wrong, but can you please help me understand where is it I am going wrong?




Re: There are 6 people at a party sitting at a round table with 6 seats: A &nbs
[#permalink]
24 Aug 2018, 13:19






