Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 24 Apr 2015, 22:36

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Intern
Joined: 21 Nov 2007
Posts: 11
Followers: 0

Kudos [?]: 3 [0], given: 0

At a dinner party , 5 people are to be seated around a [#permalink]  08 Jan 2008, 05:53
At a dinner party , 5 people are to be seated around a circular table . Two seating arrangments 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 arrangments for the group?
A. 5
B. 10
C. 24
D. 32
E. 120
Director
Joined: 12 Jul 2007
Posts: 865
Followers: 12

Kudos [?]: 217 [0], given: 0

Re: seating problem [#permalink]  08 Jan 2008, 06:00

5! = 120, but since it's a circular table we need to fix one person (5-1)! = 4! = 24. Same technique applies for all permutations where one person needs to be fixed.
Intern
Joined: 21 Nov 2007
Posts: 11
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: seating problem [#permalink]  09 Jan 2008, 01:02
eschn3am wrote:

5! = 120, but since it's a circular table we need to fix one person (5-1)! = 4! = 24. Same technique applies for all permutations where one person needs to be fixed.

Also can you please advise to me the level of this sort of problem in GMAT ?
Director
Joined: 09 Aug 2006
Posts: 766
Followers: 1

Kudos [?]: 68 [0], given: 0

Re: seating problem [#permalink]  09 Jan 2008, 01:06
Richard Lee wrote:
At a dinner party , 5 people are to be seated around a circular table . Two seating arrangments 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 arrangments for the group?
A. 5
B. 10
C. 24
D. 32
E. 120

C.

For circular arrangement, consider one place fixed and find the arrangement for the remaining seats, in this case 4! = 24
Director
Joined: 12 Jul 2007
Posts: 865
Followers: 12

Kudos [?]: 217 [0], given: 0

Re: seating problem [#permalink]  09 Jan 2008, 03:35
Richard Lee wrote:
eschn3am wrote:

5! = 120, but since it's a circular table we need to fix one person (5-1)! = 4! = 24. Same technique applies for all permutations where one person needs to be fixed.

Also can you please advise to me the level of this sort of problem in GMAT ?

I haven't taken the test before, so I really don't know. I believe that permutation and combination problems are pretty high level though
CEO
Joined: 29 Mar 2007
Posts: 2591
Followers: 16

Kudos [?]: 233 [0], given: 0

Re: seating problem [#permalink]  09 Jan 2008, 19:52
Richard Lee wrote:
At a dinner party , 5 people are to be seated around a circular table . Two seating arrangments 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 arrangments for the group?
A. 5
B. 10
C. 24
D. 32
E. 120

24. We have to fix one person so its actually 4! I'm not really sure why we have to fix one person. I just know we do. Some more insight on the reasoning behind that would be awsome.
Manager
Joined: 22 Oct 2007
Posts: 120
Followers: 1

Kudos [?]: 33 [0], given: 0

Re: seating problem [#permalink]  13 Jan 2008, 13:20
GMATBLACKBELT wrote:
Richard Lee wrote:
At a dinner party , 5 people are to be seated around a circular table . Two seating arrangments 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 arrangments for the group?
A. 5
B. 10
C. 24
D. 32
E. 120

24. We have to fix one person so its actually 4! I'm not really sure why we have to fix one person. I just know we do. Some more insight on the reasoning behind that would be awsome.

Consider 5 points chairs on a straight line (A..E) each with one person seated on it. Push each person to the seat next to him. The last one (E) will move to seat A which changes the arrangement (as the relative position of E changes).

Now consider the same 5 chairs and people on a circle. In this case E already has A as his neighbor. moving a person to the chair next to him does not change the arrangement. E still has A as his neighbor and there is no change in the relative position of A and E. Same is the case with rest of the people. A (and similarly B,C,D and E) can occupy one of the 5 chairs without changing their relative position and hence the arrangement for any particular seating sequence. Therefore we need to divide 5! by 5 to arrive at the final answer.
Hope this helps and apologies if it sounds convoluted.
Manager
Joined: 04 Jan 2008
Posts: 85
Followers: 1

Kudos [?]: 10 [0], given: 0

Re: seating problem [#permalink]  13 Jan 2008, 18:10
4! is wonderful since we can just use it as a formula. dont waste time thinking.
Manager
Joined: 01 Sep 2007
Posts: 103
Location: Astana
Followers: 1

Kudos [?]: 19 [0], given: 0

Re: seating problem [#permalink]  08 Mar 2008, 21:19
Therefore we need to divide 5! by 5 to arrive at the final answer.

I am still having troubles understanding this. Initially When I saw the problem I thought it should be 5!-5. where am i wrong? Thanks!
CEO
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
Followers: 9

Kudos [?]: 347 [1] , given: 4

Re: seating problem [#permalink]  08 Mar 2008, 21:58
1
KUDOS
CaspAreaGuy wrote:
Therefore we need to divide 5! by 5 to arrive at the final answer.

I am still having troubles understanding this. Initially When I saw the problem I thought it should be 5!-5. where am i wrong? Thanks!

http://mathworld.wolfram.com/CircularPermutation.html
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Re: seating problem   [#permalink] 08 Mar 2008, 21:58
Similar topics Replies Last post
Similar
Topics:
6 At a dinner party, 5 people are to be seated around a 11 17 Mar 2010, 06:01
9 At a dinner party, 5 people are to be seated around a circul 7 30 Oct 2009, 11:26
32 At a dinner party, 5 people are to be seated around a 12 26 Dec 2007, 07:16
At a dinner party, 5 people are to be seated around a 4 05 Dec 2006, 16:53
At a dinner party, 5 people are to be seated around a 1 22 May 2006, 03:22
Display posts from previous: Sort by