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

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At a dinner party 5 people are to be seated around a circula [#permalink] New post 23 Mar 2013, 20:36
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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
[Reveal] Spoiler: OA
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Re: At a dinner party 5 people are to be seated around a circula [#permalink] New post 23 Mar 2013, 21:00
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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 circula [#permalink] New post 24 Mar 2013, 01:21
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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:
seven-men-and-seven-women-have-to-sit-around-a-circular-92402.html
a-group-of-four-women-and-three-men-have-tickets-for-seven-a-88604.html
the-number-of-ways-in-which-5-men-and-6-women-can-be-seated-94915.html
in-how-many-different-ways-can-4-ladies-and-4-gentlemen-be-102187.html
4-couples-are-seating-at-a-round-tables-how-many-ways-can-131048.html
at-a-party-5-people-are-to-be-seated-around-a-circular-104101.html
seven-family-members-are-seated-around-their-circular-dinner-102184.html
seven-men-and-seven-women-have-to-sit-around-a-circular-11473.html
a-group-of-8-friends-sit-together-in-a-circle-alice-betty-106928.html
seven-men-and-five-women-have-to-sit-around-a-circular-table-98185.html
a-group-of-8-friends-sit-together-in-a-circle-alice-betty-106928.html
find-the-number-of-ways-in-which-four-men-two-women-and-a-106919.html
gmat-club-monday-giveaway-155157.html (700+)

Hope it helps.
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Re: At a dinner party 5 people are to be seated around a circula [#permalink] New post 24 Mar 2013, 01:58
I am a little weak in combinatorics , could some one explain to be why the answer is not 5! .
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Re: At a dinner party 5 people are to be seated around a circula [#permalink] New post 24 Mar 2013, 02:00
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shelrod007 wrote:
I am a little weak in combinatorics , could some one explain to be why the answer is not 5! .


Check here: at-a-dinner-party-5-people-are-to-be-seated-around-a-circula-149709.html#p1201656

Also, check the links there.
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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: At a dinner party 5 people are to be seated around a circula [#permalink] New post 24 Mar 2013, 07:55
HerrGrau wrote:
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.



Thank you so much HG!

This was a very simple and concise explanation! SO if there were 10 people seated at the circular table, we would solve it as 10!/10?

Thanks once again!
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Re: At a dinner party 5 people are to be seated around a circula [#permalink] New post 24 Mar 2013, 11:34
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You are very welcome. Yes, 10!/10 is exactly right for 10 people around a circular table.

HG.
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Re: At a dinner party 5 people are to be seated around a circula [#permalink] New post 24 Mar 2013, 21:01
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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"
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Re: At a dinner party 5 people are to be seated around a circula [#permalink] New post 25 Mar 2013, 02:03
shelrod007 wrote:
I am a little weak in combinatorics , could some one explain to be why the answer is not 5! .

The short answer would be something like this to your question rephrased (why isn't it n! instead of n-1! ?)

The reason is that RELATIVE to each other, ie (BAC) (CAB) ie BA AB CA AC, are seated next to each other and can be considered 'one group'
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Re: At a dinner party 5 people are to be seated around a circula [#permalink] New post 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?
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Re: At a dinner party 5 people are to be seated around a circula [#permalink] New post 23 Apr 2014, 19:18
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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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Re: At a dinner party 5 people are to be seated around a circula   [#permalink] 23 Apr 2014, 19:18
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