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• ### $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### FREE Quant Workshop by e-GMAT! December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. # At a dinner party, 5 people are to be seated around a circular table.  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags VP Joined: 22 Nov 2007 Posts: 1044 At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags Updated on: 01 Nov 2018, 01:59 5 22 00:00 Difficulty: 5% (low) Question Stats: 72% (00:23) correct 28% (00:34) wrong based on 518 sessions ### HideShow timer Statistics 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 Originally posted by marcodonzelli on 26 Dec 2007, 07:16. Last edited by Bunuel on 01 Nov 2018, 01:59, edited 2 times in total. Renamed the topic and edited the question. ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51218 At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 24 Mar 2013, 01:21 3 7 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 $$(n-1)!$$. 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: $$R = \frac{n!}{n} = (n-1)!$$" $$(n-1)!=(5-1)!=24$$ Answer: C. Questions about this concept to practice: http://gmatclub.com/forum/seven-men-and ... 92402.html http://gmatclub.com/forum/a-group-of-fo ... 88604.html http://gmatclub.com/forum/the-number-of ... 94915.html http://gmatclub.com/forum/in-how-many-d ... 02187.html http://gmatclub.com/forum/4-couples-are ... 31048.html http://gmatclub.com/forum/at-a-party-5- ... 04101.html http://gmatclub.com/forum/seven-family- ... 02184.html http://gmatclub.com/forum/seven-men-and ... 11473.html http://gmatclub.com/forum/a-group-of-8- ... 06928.html http://gmatclub.com/forum/seven-men-and ... 98185.html http://gmatclub.com/forum/a-group-of-8- ... 06928.html http://gmatclub.com/forum/find-the-numb ... 06919.html Hope it helps. _________________ ##### Most Helpful Community Reply Intern Joined: 27 Dec 2007 Posts: 48 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 07 Jan 2008, 01:24 25 11 sort cut for any circular seating arrangements: (n-1)! =(5-1)! = 4! = 4*3*2*1 = 24 ##### General Discussion CEO Joined: 17 Nov 2007 Posts: 3440 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 26 Dec 2007, 07:33 2 2 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 VP Joined: 22 Nov 2007 Posts: 1044 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 26 Dec 2007, 09:49 1 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! CEO Joined: 17 Nov 2007 Posts: 3440 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 26 Dec 2007, 10:08 6 1 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 Manager Joined: 27 Oct 2008 Posts: 177 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 27 Sep 2009, 10:48 3 1 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. Manager Joined: 10 Sep 2010 Posts: 119 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 05 Nov 2010, 20:11 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 $$(n-1)!$$. 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: $$R = \frac{n!}{n} = (n-1)!$$" $$(n-1)!=(5-1)!=24$$ Answer: C. Similar question: combinatrics-86547.html?hilit=relative%20around Hope 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 (n-1)!, 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? Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 05 Nov 2010, 20:40 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 $$(n-1)!$$. 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: $$R = \frac{n!}{n} = (n-1)!$$" $$(n-1)!=(5-1)!=24$$ Answer: C. Similar question: combinatrics-86547.html?hilit=relative%20around Hope 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 (n-1)!, 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 $$(n-1)!=4!=24$$, (120 would be the answer if they were arranged in a row). _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8678 Location: Pune, India Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 06 Nov 2010, 04:08 1 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. _________________ 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: 27 Jan 2013 Posts: 228 GMAT 1: 780 Q49 V51 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 23 Mar 2013, 21:00 3 4 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. _________________ "It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land GMAT vs GRE Comparison If you found my post useful KUDOS are much appreciated. IMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE: Here is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8678 Location: Pune, India Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 24 Mar 2013, 21:01 2 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 (n-1)!) 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" _________________ 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: 15 Aug 2013 Posts: 247 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 23 Apr 2014, 18:41 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 (n-1)!) 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? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8678 Location: Pune, India Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 23 Apr 2014, 19:18 1 1 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 ... ts-part-i/ http://www.veritasprep.com/blog/2011/11 ... 3-part-ii/ _________________ 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: 21 Feb 2015 Posts: 10 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 22 Feb 2015, 09:43 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. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13095 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: At a dinner party, 5 people are to be seated around a circular table. [#permalink] ### Show Tags 22 Feb 2015, 14:00 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 'set-up' 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: At a dinner party, 5 people are to be seated around a circular table.  [#permalink]

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01 Mar 2018, 17:37
1
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.

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Re: At a dinner party, 5 people are to be seated around a circular table.  [#permalink]

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11 Dec 2018, 03:08
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 $$(n-1)!$$.

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:

$$R = \frac{n!}{n} = (n-1)!$$"

$$(n-1)!=(5-1)!=24$$

http://gmatclub.com/forum/seven-men-and ... 92402.html
http://gmatclub.com/forum/a-group-of-fo ... 88604.html
http://gmatclub.com/forum/the-number-of ... 94915.html
http://gmatclub.com/forum/in-how-many-d ... 02187.html
http://gmatclub.com/forum/4-couples-are ... 31048.html
http://gmatclub.com/forum/at-a-party-5- ... 04101.html
http://gmatclub.com/forum/seven-family- ... 02184.html
http://gmatclub.com/forum/seven-men-and ... 11473.html
http://gmatclub.com/forum/a-group-of-8- ... 06928.html
http://gmatclub.com/forum/seven-men-and ... 98185.html
http://gmatclub.com/forum/a-group-of-8- ... 06928.html
http://gmatclub.com/forum/find-the-numb ... 06919.html
http://gmatclub.com/forum/gmat-club-mon ... 55157.html (700+)

Hope it helps.

Bunuel,

Thanks.
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Posts: 51218
Re: At a dinner party, 5 people are to be seated around a circular table.  [#permalink]

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11 Dec 2018, 03:10
1
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 $$(n-1)!$$.

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:

$$R = \frac{n!}{n} = (n-1)!$$"

$$(n-1)!=(5-1)!=24$$

http://gmatclub.com/forum/seven-men-and ... 92402.html
http://gmatclub.com/forum/a-group-of-fo ... 88604.html
http://gmatclub.com/forum/the-number-of ... 94915.html
http://gmatclub.com/forum/in-how-many-d ... 02187.html
http://gmatclub.com/forum/4-couples-are ... 31048.html
http://gmatclub.com/forum/at-a-party-5- ... 04101.html
http://gmatclub.com/forum/seven-family- ... 02184.html
http://gmatclub.com/forum/seven-men-and ... 11473.html
http://gmatclub.com/forum/a-group-of-8- ... 06928.html
http://gmatclub.com/forum/seven-men-and ... 98185.html
http://gmatclub.com/forum/a-group-of-8- ... 06928.html
http://gmatclub.com/forum/find-the-numb ... 06919.html
http://gmatclub.com/forum/gmat-club-mon ... 55157.html (700+)

Hope it helps.

Bunuel,

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.  [#permalink]

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11 Dec 2018, 04:08
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 ( n-1)!

so ( 5-1)! ;4! = 24 IMO C
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Re: At a dinner party, 5 people are to be seated around a circular table. &nbs [#permalink] 11 Dec 2018, 04:08
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