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

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VP
Joined: 22 Nov 2007
Posts: 1063
At a dinner party, 5 people are to be seated around a  [#permalink]

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Updated on: 08 Jul 2013, 07:37
5
15
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Difficulty:

5% (low)

Question Stats:

74% (00:14) correct 26% (00:20) wrong based on 740 sessions

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At a dinner party, 5 people are to be seated around a circular table. 2 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: at-a-dinner-party-5-people-are-to-be-seated-around-a-circula-149709.html

Originally posted by marcodonzelli on 26 Dec 2007, 08:16.
Last edited by Bunuel on 08 Jul 2013, 07:37, edited 1 time in total.
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26 Dec 2007, 08:33
1
2
C

let 1,2,3,4,5 are people.

1. we fix position of 1
2. we have 4*3=12 possible positions for left and right neighbors of 1.
3. for each position of x1y we have 2 possible positions for last two people: ax1yb and bx1ya.

Therefore, N=12*2=24
VP
Joined: 22 Nov 2007
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26 Dec 2007, 10:49
1
OA is C, good...anyway, provided I am quite bad at combs, would you please explain it to me step by step in a very clear way..i cannot catch your 3 points!
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26 Dec 2007, 11:08
6
1
marcodonzelli wrote:
OA is C, good...anyway, provided I am quite bad at combs, would you please explain it to me step by step in a very clear way..i cannot catch your 3 points!

let 1,2,3,4,5 are people.
"_ _ _ _ _" - positions

1. we fix position of 1
"_ _ 1 _ _"

2. we have 4*3=12 possible positions for left and right neighbors of 1.
"_ x 1 _ _" x e {2,3,4,5}. 4 variants
"_ x 1 y _" y e {(2,3,4,5} - {x}. 3 variants

total number of variants is 4*3=12

3. for each position of x1y we have 2 possible positions for last two people: ax1yb and bx1ya.
or
"a x 1 y _" a e {(2,3,4,5} - {x,y}. 2 variants
"a x 1 y b" b e {(2,3,4,5} - {x,y,a}. 1 variants

Therefore, N=12*2=24
Manager
Joined: 27 Dec 2007
Posts: 55

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07 Jan 2008, 02:24
25
11
sort cut for any circular seating arrangements:
(n-1)! =(5-1)! = 4! = 4*3*2*1 = 24
Manager
Joined: 27 Oct 2008
Posts: 180

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27 Sep 2009, 11:48
3
1
At a dinner party, 5 people are to be seated around a circular table. 2 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

Soln: Since the arrangement is circular and 2 seating arrangements are considered different only when the positions of the people are different relative to each other, we can find the total number of possible seating arrangements, by fixing one person's position and arranging the others.

Thus if one person's position is fixed, the others can be arranged in 4! ways.
Ans is C.
Intern
Joined: 25 Jan 2010
Posts: 5

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26 Jan 2010, 23:54
2
At first lets see, there are how many cases to sit, if the they will sit not in round table.
there are 5*4*3*2*1=5! cases
And because it is no important that which sit is first that is mean that we must devise this to 5,
prob=5!/5=4!=24
Intern
Joined: 19 Dec 2009
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27 Jan 2010, 02:42
1
The solution Araik mentioned is logical as this was a problem with a round table. As a rule one needs to understand that there is a difference is arrangement in a row and that of a circle. Please refer to Math Workbook for 'COMBINATORICS'. Its explained very well there.
Manager
Joined: 27 Jul 2010
Posts: 171
Location: Prague
Schools: University of Economics Prague

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03 Feb 2011, 05:54
1
Is not this a normal circular permutation problem?

The solution: (5-1)!
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27 Aug 2011, 06:05
1
navami wrote:
sort cut for any circular seating arrangements:
(n-1)! =(5-1)! = 4! = 4*3*2*1 = 24

hi, kinda skeptical about the shortcut...do you really mean that when we see any circular table question, we can just plug in the (n-1)! formula?

So if it is a 8 table round table sitting arrangement, I can just use (8-1)! to get question resolved? Any other criteria or...does it limit only to certain criteria? Pls enlighten...many thanks!
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27 Aug 2011, 06:08
1
miweekend wrote:
navami wrote:
sort cut for any circular seating arrangements:
(n-1)! =(5-1)! = 4! = 4*3*2*1 = 24

hi, kinda skeptical about the shortcut...do you really mean that when we see any circular table question, we can just plug in the (n-1)! formula?

So if it is a 8 table round table sitting arrangement, I can just use (8-1)! to get question resolved? Any other criteria or...does it limit only to certain criteria? Pls enlighten...many thanks!

(n-1)! is correct for this constraint free question irrespective of the number of people in question. However, you may have to apply some logic if there are other constraints, such as A can't sit with B, OR D and E must sit together.
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Re: At a dinner party, 5 people are to be seated around a  [#permalink]

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08 Jul 2013, 07:24
3
At a dinner party, 5 people are to be seated around a circular table. 2 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?

If the people sit in a row instead of a circular table, we would have 5*4*3*2*1 = 120 different combinations.
But (restriction), if the people spin around the table without being in different position relative to each other does not generates a new position.
Then, we have to eliminate ABCDE, BCDEA, CDEAB, DEABC and EABCD. Total: 5 arrangements.
Then, instead of 5*4*3*2*1 = 120, we have to eliminate the 5 arrangements --> 5*4*3*2*1 = 24

After reading the solution given by walker, I thought in this one:

Fixing 1 person in the table (C for example), and moving the neighbours of this person.

Slots: 12345 --> We fix the third slot with person "C" --> 12C34--> then we have four possibilities for the slot 1 (person A, B, D or E), three possibilities for the second slot (ABDE less the person that we placed in the slot 1), two for the fourth slot and one for the last slot:

4*3*1*2*1 = 24
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Joined: 02 Sep 2009
Posts: 47920
Re: At a dinner party, 5 people are to be seated around a  [#permalink]

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08 Jul 2013, 07:41
14
15
Maxirosario2012 wrote:
At a dinner party, 5 people are to be seated around a circular table. 2 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?

If the people sit in a row instead of a circular table, we would have 5*4*3*2*1 = 120 different combinations.
But (restriction), if the people spin around the table without being in different position relative to each other does not generates a new position.
Then, we have to eliminate ABCDE, BCDEA, CDEAB, DEABC and EABCD. Total: 5 arrangements.
Then, instead of 5*4*3*2*1 = 120, we have to eliminate the 5 arrangements --> 5*4*3*2*1 = 24

After reading the solution given by walker, I thought in this one:

Fixing 1 person in the table (C for example), and moving the neighbours of this person.

Slots: 12345 --> We fix the third slot with person "C" --> 12C34--> then we have four possibilities for the slot 1 (person A, B, D or E), three possibilities for the second slot (ABDE less the person that we placed in the slot 1), two for the fourth slot and one for the last slot:

4*3*1*2*1 = 24

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: at-a-dinner-party-5-people-are-to-be-seated-around-a-circula-149709.html
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Re: At a dinner party, 5 people are to be seated around a  [#permalink]

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