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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] New post 01 Nov 2010, 21:52
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A
B
C
D
E

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80% (01:21) correct 20% (01:03) wrong based on 64 sessions
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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Re: combinations problem [#permalink] New post 01 Nov 2010, 21:56
Expert's post
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.
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Re: combinations problem [#permalink] New post 05 Nov 2010, 12:37
+1 C 8-)

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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Re: combinations problem [#permalink] New post 05 Nov 2010, 19:53
Expert's post
metallicafan wrote:
+1 C 8-)

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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Re: combinations problem [#permalink] New post 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?
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Re: combinations problem [#permalink] New post 05 Nov 2010, 20:40
Expert's post
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).
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: combinations problem [#permalink] New post 06 Nov 2010, 04:08
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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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need solution to this question [#permalink] New post 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
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Re: need solution to this question [#permalink] New post 12 Dec 2010, 00:39
Expert's post
Merging similar topics.

Questions about the same concept:
arrangements-around-the-table-102184.html?
ways-to-sit-around-the-table-102187.html?
please-help-with-the-seating-arrangement-problems-94915.html?
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: combinations problem [#permalink] New post 13 Dec 2010, 06:30
Thanks for that formula.

That really helps
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Re: combinations problem   [#permalink] 13 Dec 2010, 06:30
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