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At a dinner party, 5 people are to be seated around a circular table.
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At a dinner party, 5 people are to be seated around a circular table. 2 seating arrangements are considered different only when the positions of the people are different relative to each other. what is the total number of different possible seating arrangements for the group? A. 5 B. 10 C. 24 D. 32 E. 120
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Originally posted by marcodonzelli on 26 Dec 2007, 08:16.
Last edited by Bunuel on 01 Nov 2018, 02:59, edited 2 times in total.
Renamed the topic and edited the question.




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At a dinner party, 5 people are to be seated around a circular table.
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24 Mar 2013, 02:21
Val1986 wrote: At a dinner party 5 people are to be seated around a circular table. Two sitting arrangements are considered different only when the positions of the people are different relative to each other.What is the total number of possible sitting arrangements or the group?
A. 5 B. 10 C. 24 D. 32 E. 120 We have a case of circular arrangement. The number of arrangements of n distinct objects in a row is given by \(n!\). The number of arrangements of n distinct objects in a circle is given by \((n1)!\). From Gmat Club Math Book (combinatorics chapter): "The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have: \(R = \frac{n!}{n} = (n1)!\)" \((n1)!=(51)!=24\) Answer: C. Questions about this concept to practice: http://gmatclub.com/forum/sevenmenand ... 92402.htmlhttp://gmatclub.com/forum/agroupoffo ... 88604.htmlhttp://gmatclub.com/forum/thenumberof ... 94915.htmlhttp://gmatclub.com/forum/inhowmanyd ... 02187.htmlhttp://gmatclub.com/forum/4couplesare ... 31048.htmlhttp://gmatclub.com/forum/ataparty5 ... 04101.htmlhttp://gmatclub.com/forum/sevenfamily ... 02184.htmlhttp://gmatclub.com/forum/sevenmenand ... 11473.htmlhttp://gmatclub.com/forum/agroupof8 ... 06928.htmlhttp://gmatclub.com/forum/sevenmenand ... 98185.htmlhttp://gmatclub.com/forum/agroupof8 ... 06928.htmlhttp://gmatclub.com/forum/findthenumb ... 06919.htmlHope it helps.
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Re: At a dinner party, 5 people are to be seated around a circular table.
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07 Jan 2008, 02:24
sort cut for any circular seating arrangements: (n1)! =(51)! = 4! = 4*3*2*1 = 24




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Re: At a dinner party, 5 people are to be seated around a circular table.
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26 Dec 2007, 08:33
C
let 1,2,3,4,5 are people.
1. we fix position of 1
2. we have 4*3=12 possible positions for left and right neighbors of 1.
3. for each position of x1y we have 2 possible positions for last two people: ax1yb and bx1ya.
Therefore, N=12*2=24



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Re: At a dinner party, 5 people are to be seated around a circular table.
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26 Dec 2007, 10:49
OA is C, good...anyway, provided I am quite bad at combs, would you please explain it to me step by step in a very clear way..i cannot catch your 3 points!



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Re: At a dinner party, 5 people are to be seated around a circular table.
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26 Dec 2007, 11:08
marcodonzelli wrote: OA is C, good...anyway, provided I am quite bad at combs, would you please explain it to me step by step in a very clear way..i cannot catch your 3 points!
let 1,2,3,4,5 are people.
"_ _ _ _ _"  positions
1. we fix position of 1
"_ _ 1 _ _"
2. we have 4*3=12 possible positions for left and right neighbors of 1.
"_ x 1 _ _" x e {2,3,4,5}. 4 variants "_ x 1 y _" y e {(2,3,4,5}  {x}. 3 variants
total number of variants is 4*3=12
3. for each position of x1y we have 2 possible positions for last two people: ax1yb and bx1ya.
or "a x 1 y _" a e {(2,3,4,5}  {x,y}. 2 variants "a x 1 y b" b e {(2,3,4,5}  {x,y,a}. 1 variants
Therefore, N=12*2=24



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Re: At a dinner party, 5 people are to be seated around a circular table.
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27 Sep 2009, 11:48
At a dinner party, 5 people are to be seated around a circular table. 2 seating arrangements are considered different only when the positions of the people are different relative to each other. what is the total number of different possible seating arrangements for the group?
A. 5 B. 10 C. 24 D. 32 E. 120
Soln: Since the arrangement is circular and 2 seating arrangements are considered different only when the positions of the people are different relative to each other, we can find the total number of possible seating arrangements, by fixing one person's position and arranging the others.
Thus if one person's position is fixed, the others can be arranged in 4! ways. Ans is C.



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Re: At a dinner party, 5 people are to be seated around a circular table.
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05 Nov 2010, 21:11
Bunuel wrote: mybudgie wrote: At a party, 5 people are to be seated around a circular table; two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements? A 5 B 10 C 24 D 32 E 120 This is the case of circular arrangement. The number of arrangements of n distinct objects in a row is given by \(n!\). The number of arrangements of n distinct objects in a circle is given by \((n1)!\). From Gmat Club Math Book (combinatorics chapter): "The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have: \(R = \frac{n!}{n} = (n1)!\)" \((n1)!=(51)!=24\) Answer: C. Similar question: combinatrics86547.html?hilit=relative%20aroundHope it's clear. I was confused by the wording of the question "only when the positions of the people are different relative to each other". I knew the formula (n1)!, but I though that the correct answer would require some limited set of the combinations, defined by "only when the positions of the people are different relative to each other". Then if the question did not mentioned this special condition of ""only when the positions of the people are different relative to each other", the answer would be 24x5=120. Right?



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Re: At a dinner party, 5 people are to be seated around a circular table.
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05 Nov 2010, 21:40
Fijisurf wrote: Bunuel wrote: mybudgie wrote: At a party, 5 people are to be seated around a circular table; two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements? A 5 B 10 C 24 D 32 E 120 This is the case of circular arrangement. The number of arrangements of n distinct objects in a row is given by \(n!\). The number of arrangements of n distinct objects in a circle is given by \((n1)!\). From Gmat Club Math Book (combinatorics chapter): "The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have: \(R = \frac{n!}{n} = (n1)!\)" \((n1)!=(51)!=24\) Answer: C. Similar question: combinatrics86547.html?hilit=relative%20aroundHope it's clear. I was confused by the wording of the question "only when the positions of the people are different relative to each other". I knew the formula (n1)!, but I though that the correct answer would require some limited set of the combinations, defined by "only when the positions of the people are different relative to each other". Then if the question did not mentioned this special condition of ""only when the positions of the people are different relative to each other", the answer would be 24x5=120. Right? "the positions of the people are different relative to each other" just means different arrangements (around a circular table). The number of arrangements of n distinct objects in a circle is \((n1)!=4!=24\), (120 would be the answer if they were arranged in a row).
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Re: At a dinner party, 5 people are to be seated around a circular table.
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06 Nov 2010, 05:08
Arranging 3 people (A, B, C) in a row: A B C, A C B, B A C. B C A, C A B, C B A 3! ways Why is arranging 3 people in a circle different? ....A ....O B......C If I am B, A is to my left, C is to my right. Look at this one now: ....C ....O A......B Here also, A is to my left and C is to my right. In a circle, these are considered a single arrangement because relative to each other, people are still sitting in the same position. This is the general rule in circular arrangement. You use the formula n!/n = (n  1)! because every n arrangements are considered a single arrangement. e.g. if n = 3, the given 3 arrangements are the same: .....A ................ C ............... B .....O ................ O .............. O B........C ........ A ..... B ..... C........ A In each of these, if I am B, I am sitting in the same position relative to others. A is to my left and C is to my right. and these three are the same: .....C ................ A ............... B .....O ................ O .............. O B........A ........ C ..... B ..... A........ C Here, if I am B, C is to my left and A is to my right. Different from the first three. Hence no. of arrangements = 3!/3 = 2 only Here, they have mentioned 'relative to people' only to make it clearer. In a circle, anyway only relative to people arrangements are considered. You might need to use n! in a circle if they mention that each seat in the circular arrangement is numbered and is hence different etc. Then there are just n distinct seats and n people. If nothing of the sorts is mentioned, you always use the (n  1)! formula for circular arrangement.
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Re: At a dinner party, 5 people are to be seated around a circular table.
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23 Mar 2013, 22:00
Hi there, You can treat this as an ordering question except that for a circular arrangement you need to divide by the number of spaces. So in this case: 5!/5=24 If you spin the circle to right, that doesn't count as a new arrangement. Dividing by the number of spaces takes that into consideration. Happy Studies, HG.
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Re: At a dinner party, 5 people are to be seated around a circular table.
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24 Mar 2013, 22:01
Val1986 wrote: At a dinner party 5 people are to be seated around a circular table. Two sitting arrangements are considered different only when the positions of the people are different relative to each other.What is the total number of possible sitting arrangements or the group?
A. 5 B. 10 C. 24 D. 32 E. 120 Check out this post on circular arrangements. It discusses why the number of arrangements is n!/n (which is the same as (n1)!) in case there are n people sitting around a round table. http://www.veritasprep.com/blog/2011/10 ... angements/It also discusses the relevance of this statement in the question: "Two sitting arrangements are considered different only when the positions of the people are different relative to each other"
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Re: At a dinner party, 5 people are to be seated around a circular table.
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23 Apr 2014, 19:41
VeritasPrepKarishma wrote: Val1986 wrote: At a dinner party 5 people are to be seated around a circular table. Two sitting arrangements are considered different only when the positions of the people are different relative to each other.What is the total number of possible sitting arrangements or the group?
A. 5 B. 10 C. 24 D. 32 E. 120 Check out this post on circular arrangements. It discusses why the number of arrangements is n!/n (which is the same as (n1)!) in case there are n people sitting around a round table. http://www.veritasprep.com/blog/2011/10 ... angements/It also discusses the relevance of this statement in the question: "Two sitting arrangements are considered different only when the positions of the people are different relative to each other" Hi Karishma, If there were constraints such as A can't be next to B or C, does that mean that we now have 5 seats but since 3 of them are fixed, the solution would be 2!/2?



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Re: At a dinner party, 5 people are to be seated around a circular table.
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23 Apr 2014, 20:18
russ9 wrote: Hi Karishma,
If there were constraints such as A can't be next to B or C, does that mean that we now have 5 seats but since 3 of them are fixed, the solution would be 2!/2?
I am assuming your question is this: 5 people are to be seated around a circular table such that A sits neither next to B nor next to C. How many arrangements are possible? I don't know how you consider "...3 of them are fixed". The way you handle this constraint would be this: There are 5 vacant seats. Make A occupy 1 seat in 1 way (because all seats are same before anybody sits). Now we have 4 unique vacant seats (unique with respect to A) and 4 people. B and C cannot sit next to A so D and E occupy the seats right next to A on either side. This can be done in 2! ways: D A E or E A D B and C occupy the two unique seats away from A. This can be done in 2! ways. Total number of arrangements = 2! * 2! = 4 Check out these posts. First discusses theory of circular arrangements and next two discuss circular arrangements with various constraints: http://www.veritasprep.com/blog/2011/10 ... angements/http://www.veritasprep.com/blog/2011/10 ... tsparti/http://www.veritasprep.com/blog/2011/11 ... 3partii/
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Re: At a dinner party, 5 people are to be seated around a circular table.
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22 Feb 2015, 10:43
Please provide feedback to see if this makes sense:
I approached it thinking, if you have a circle, say, ABCDE there would be 5! ways of arranging, however, the question states that an arrangement is only different if the positions relative to each other are different.
So (1/5)th of the time each person in the circle would in the same position relative to another person. Therefore I did (1/5)x5! = 24
I was thinking "symmetry" as well  what are the experts thoughts on this? The formula though is definitely the easier way.



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Re: At a dinner party, 5 people are to be seated around a circular table.
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22 Feb 2015, 15:00
Hi icetray, Your thinking on this question is just fine. Conceptually, since we're dealing with a circular table with 5 chairs (and not a row of chairs), the table could have 5 different "starting chairs." As such the arrangements (going around the table): ABCDE BCDEA CDEAB DEABC EABCD Are all the same arrangement (just 'revolved' around the table). Since we're NOT allowed to count each of those (they're not different arrangements, they're just rotations of the same arrangement), we have to divide the permutation by 5. 5!/5 = 24 This type of 'setup' is relatively rare on Test Day  there's a pretty good chance that you won't see it at all. If you do though, then your way of handling the "math" is just as viable as the formula that was given. GMAT assassins aren't born, they're made, Rich
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Re: At a dinner party, 5 people are to be seated around a circular table.
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01 Mar 2018, 18:37
Val1986 wrote: At a dinner party 5 people are to be seated around a circular table. Two sitting arrangements are considered different only when the positions of the people are different relative to each other.What is the total number of possible sitting arrangements or the group?
A. 5 B. 10 C. 24 D. 32 E. 120 When determining the number way to arrange a group around a circle, we subtract 1 from the total and set it to a factorial. Thus, the total number of possible sitting arrangements for 5 people around a circular table is 4! = 24. Answer: C
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Re: At a dinner party, 5 people are to be seated around a circular table.
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11 Dec 2018, 04:08
Bunuel wrote: Val1986 wrote: At a dinner party 5 people are to be seated around a circular table. Two sitting arrangements are considered different only when the positions of the people are different relative to each other.What is the total number of possible sitting arrangements or the group?
A. 5 B. 10 C. 24 D. 32 E. 120 We have a case of circular arrangement. The number of arrangements of n distinct objects in a row is given by \(n!\). The number of arrangements of n distinct objects in a circle is given by \((n1)!\). From Gmat Club Math Book (combinatorics chapter): "The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have: \(R = \frac{n!}{n} = (n1)!\)" \((n1)!=(51)!=24\) Answer: C. Questions about this concept to practice: http://gmatclub.com/forum/sevenmenand ... 92402.htmlhttp://gmatclub.com/forum/agroupoffo ... 88604.htmlhttp://gmatclub.com/forum/thenumberof ... 94915.htmlhttp://gmatclub.com/forum/inhowmanyd ... 02187.htmlhttp://gmatclub.com/forum/4couplesare ... 31048.htmlhttp://gmatclub.com/forum/ataparty5 ... 04101.htmlhttp://gmatclub.com/forum/sevenfamily ... 02184.htmlhttp://gmatclub.com/forum/sevenmenand ... 11473.htmlhttp://gmatclub.com/forum/agroupof8 ... 06928.htmlhttp://gmatclub.com/forum/sevenmenand ... 98185.htmlhttp://gmatclub.com/forum/agroupof8 ... 06928.htmlhttp://gmatclub.com/forum/findthenumb ... 06919.htmlhttp://gmatclub.com/forum/gmatclubmon ... 55157.html (700+) Hope it helps. Bunuel, Why am I not authorized to see the contents of the last link ( http://gmatclub.com/forum/gmatclubmon ... 55157.html)? Please help me. Thanks.
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Re: At a dinner party, 5 people are to be seated around a circular table.
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11 Dec 2018, 04:10
SpiritualYoda wrote: Bunuel wrote: Val1986 wrote: At a dinner party 5 people are to be seated around a circular table. Two sitting arrangements are considered different only when the positions of the people are different relative to each other.What is the total number of possible sitting arrangements or the group?
A. 5 B. 10 C. 24 D. 32 E. 120 We have a case of circular arrangement. The number of arrangements of n distinct objects in a row is given by \(n!\). The number of arrangements of n distinct objects in a circle is given by \((n1)!\). From Gmat Club Math Book (combinatorics chapter): "The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have: \(R = \frac{n!}{n} = (n1)!\)" \((n1)!=(51)!=24\) Answer: C. Questions about this concept to practice: http://gmatclub.com/forum/sevenmenand ... 92402.htmlhttp://gmatclub.com/forum/agroupoffo ... 88604.htmlhttp://gmatclub.com/forum/thenumberof ... 94915.htmlhttp://gmatclub.com/forum/inhowmanyd ... 02187.htmlhttp://gmatclub.com/forum/4couplesare ... 31048.htmlhttp://gmatclub.com/forum/ataparty5 ... 04101.htmlhttp://gmatclub.com/forum/sevenfamily ... 02184.htmlhttp://gmatclub.com/forum/sevenmenand ... 11473.htmlhttp://gmatclub.com/forum/agroupof8 ... 06928.htmlhttp://gmatclub.com/forum/sevenmenand ... 98185.htmlhttp://gmatclub.com/forum/agroupof8 ... 06928.htmlhttp://gmatclub.com/forum/findthenumb ... 06919.htmlhttp://gmatclub.com/forum/gmatclubmon ... 55157.html (700+) Hope it helps. Bunuel, Why am I not authorized to see the contents of the last link ( http://gmatclub.com/forum/gmatclubmon ... 55157.html)? Please help me. Thanks. Ignore that question. Removed the link.
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Re: At a dinner party, 5 people are to be seated around a circular table.
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marcodonzelli wrote: At a dinner party, 5 people are to be seated around a circular table. 2 seating arrangements are considered different only when the positions of the people are different relative to each other. what is the total number of different possible seating arrangements for the group?
A. 5 B. 10 C. 24 D. 32 E. 120 for circular arrangement s ( n1)! so ( 51)! ;4! = 24 IMO C
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