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# At a dinner party 5 people are to be seated around a circula

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Intern
Joined: 20 Oct 2010
Posts: 7
At a dinner party 5 people are to be seated around a circular table.  [#permalink]

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01 Nov 2010, 22:52
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72% (00:25) correct 28% (00:39) wrong based on 397 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 47920
Re: At a dinner party 5 people are to be seated around a circula  [#permalink]

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24 Mar 2013, 02:21
2
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:

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

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

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Hope it helps.
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##### General Discussion
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Joined: 10 Sep 2010
Posts: 121
Re: At a dinner party 5 people are to be seated around a circular table.  [#permalink]

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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 $$(n-1)!$$.

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:

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

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

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: 47920
Re: At a dinner party 5 people are to be seated around a circular table.  [#permalink]

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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 $$(n-1)!$$.

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:

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

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

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

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06 Nov 2010, 05:08
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.
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Karishma
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! 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 circula [#permalink] ### Show Tags 23 Mar 2013, 22:00 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: 8184 Location: Pune, India Re: At a dinner party 5 people are to be seated around a circula [#permalink] ### Show Tags 24 Mar 2013, 22:01 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 Save up to$1,000 on GMAT prep through 8/20! Learn more here >

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Senior Manager
Joined: 15 Aug 2013
Posts: 268
Re: At a dinner party 5 people are to be seated around a circula  [#permalink]

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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 (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?
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Joined: 16 Oct 2010
Posts: 8184
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Re: At a dinner party 5 people are to be seated around a circula  [#permalink]

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23 Apr 2014, 20:18
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/
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! 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, 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. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12181 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, 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 '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 circula  [#permalink]

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01 Mar 2018, 18:37
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 circula &nbs [#permalink] 01 Mar 2018, 18:37
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