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# At a party, 5 people are to be seated around a circular

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At a party, 5 people are to be seated around a circular [#permalink]

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01 Nov 2010, 21:52
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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
[Reveal] Spoiler: OA

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01 Nov 2010, 21:56
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.
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05 Nov 2010, 12:37
+1 C

Bunuel, could you illustrate or provide more detal about this explanation? I don't understand it very well:

Bunuel wrote:
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.

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05 Nov 2010, 19:53
metallicafan wrote:
+1 C

Bunuel, could you illustrate or provide more detal about this explanation? I don't understand it very well:

Bunuel wrote:
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.

Circular arrangements: math-combinatorics-87345.html
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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:

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

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

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?

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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:

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

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

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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06 Nov 2010, 04:08
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Expert's post
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18116 [1], given: 236 Intern Joined: 13 Aug 2010 Posts: 6 Kudos [?]: [0], given: 0 need solution to this question [#permalink] ### Show Tags 12 Dec 2010, 00:30 Ques :- At a dinner 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 for the group? a. 5 b. 10 c. 24 d. 32 e. 120 Kudos [?]: [0], given: 0 Math Expert Joined: 02 Sep 2009 Posts: 42583 Kudos [?]: 135540 [0], given: 12697 Re: need solution to this question [#permalink] ### Show Tags 12 Dec 2010, 00:39 Kudos [?]: 135540 [0], given: 12697 Senior Manager Status: Bring the Rain Joined: 17 Aug 2010 Posts: 389 Kudos [?]: 47 [0], given: 46 Location: United States (MD) Concentration: Strategy, Marketing Schools: Michigan (Ross) - Class of 2014 GMAT 1: 730 Q49 V39 GPA: 3.13 WE: Corporate Finance (Aerospace and Defense) Re: combinations problem [#permalink] ### Show Tags 13 Dec 2010, 06:30 Thanks for that formula. That really helps _________________ Kudos [?]: 47 [0], given: 46 Non-Human User Joined: 09 Sep 2013 Posts: 14869 Kudos [?]: 287 [0], given: 0 Re: At a party, 5 people are to be seated around a circular [#permalink] ### Show Tags 07 Feb 2015, 10:35 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 287 [0], given: 0 Current Student Joined: 21 Feb 2015 Posts: 10 Kudos [?]: 3 [0], given: 11 Re: At a party, 5 people are to be seated around a circular [#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. Kudos [?]: 3 [0], given: 11 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 10378 Kudos [?]: 3685 [0], given: 173 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: At a party, 5 people are to be seated around a circular [#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 party, 5 people are to be seated around a circular [#permalink]

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30 Mar 2016, 13:23
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: At a party, 5 people are to be seated around a circular   [#permalink] 30 Mar 2016, 13:23
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