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

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

by BTGmoderatorDC » Thu Apr 25, 2019 7:14 pm

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A

B

C

D

E

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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

OA C

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by [email protected] » Thu Apr 25, 2019 8:35 pm
BTGmoderatorDC 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?

A. 5
B. 10
C. 24
D. 32
E. 120

OA C

Source: GMAT Prep
Note that the number of distinct arrangments of n distinguishable objects to be placed in a row = n!; while that the number of distinct arrangments of n distinguishable objects to be placed in a circle = (n - 1)!

It is because, in the case of arrangement in a circle, two seating arrangements are NOT considered different when the positions of the people are NOT different relative to each other. For example, say there are four friends A, B, C, and D. The arrangement in rows ABCD and BCDA are certainly two different; however, If A, B, C, and D are seated in a round table arrangement the arrangement in circles ABCD and BCDA are NOT two different since the positions of the people in circular arrangement ABCD and in circular arrangement BCDA are NOT different relative to each other.

Thus, the total number of different possible seating arrangements for a group of 5 people = (5 - 1)! = 4! = 24

Hope this helps!

-Jay
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by swerve » Fri Apr 26, 2019 7:22 am
Let 1,2,3,4,5 are people.

1. we fix the 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 the last two people: $$ax1yb$$ and $$bx1ya$$.

Therefore, $$N=12*2=24$$.

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by [email protected] » Mon May 06, 2019 7:13 pm
BTGmoderatorDC 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?

A. 5
B. 10
C. 24
D. 32
E. 120

OA C

Source: GMAT Prep
When determining the number of ways to arrange a group around a circle, we subtract 1 from the total and set it to a factorial. Thus, the total number of possible sitting arrangements for 5 people around a circular table is (5 - 1)! = 4! = 24.

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by [email protected] » Mon May 06, 2019 7:13 pm
[email protected] wrote:
BTGmoderatorDC 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?

A. 5
B. 10
C. 24
D. 32
E. 120

OA C
If both Sue and Jane do make the team, the number of ways to select the team is 7C3 since we need to select 3 more players from the remaining 7 players:

7C3 = 7!/(3! x 4!) = (7 x 6 x 5)/3! = 35 ways

If both Sue and Jane do not make the team, the number of ways to select the team is 7C5:

7C5 = 7!/(5! x 2!) = (7 x 6 x 5 x 4 x 3)/(5 x 4 x 3 x 2) = 7 x 3 = 21 ways

So the total possible ways is 35 + 21 = 56.

Scott Woodbury-Stewart
Founder and CEO
[email protected]

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

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by [email protected] » Mon May 06, 2019 7:13 pm
[email protected] wrote:
BTGmoderatorDC 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?

A. 5
B. 10
C. 24
D. 32
E. 120

OA C
If both Sue and Jane do make the team, the number of ways to select the team is 7C3 since we need to select 3 more players from the remaining 7 players:

7C3 = 7!/(3! x 4!) = (7 x 6 x 5)/3! = 35 ways

If both Sue and Jane do not make the team, the number of ways to select the team is 7C5:

7C5 = 7!/(5! x 2!) = (7 x 6 x 5 x 4 x 3)/(5 x 4 x 3 x 2) = 7 x 3 = 21 ways

So the total possible ways is 35 + 21 = 56.

Scott Woodbury-Stewart
Founder and CEO
[email protected]

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

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

by [email protected] » Tue Apr 12, 2022 9:29 am
BTGmoderatorDC wrote:
Thu Apr 25, 2019 7:14 pm
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

OA C

Source: GMAT Prep
Although we can quickly apply the circular arrangement formula (i.e., number of ways to arrange n objects in a circle = (n - 1)!), we can also solve the question using the Fundamental Counting Principle (FPC, aka the slot method). In the process of doing so, you'll also learn WHY the circular arrangement formula works

First label the five chairs as follows:

We can seat the first guest in one of the 5 available chairs.
We can seat the next guest in one of the 4 remaining chairs.
We can seat the next guest in one of the 3 remaining chairs.
We can seat the next guest in one of the 2 remaining chairs.
We can seat the last guest in the 1 remaining chair.
So, the total number of ways to seat the guests = (5)(4)(3)(2)(1) = 120 ways

The answer, however, is NOT E, because we have inadvertently counted every possible arrangement 5 times.

For example, the five arrangements shown here...

... are all the same, because the relative positions of the five people are the same in each case.

Since we have counted each unique arrangement 5 times, we must divide 120 by 5 to get 24 possible arrangements