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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)!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.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
Scott@TargetTestPrep wrote: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: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
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.
Answer: C
Scott@TargetTestPrep wrote: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: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
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.
Answer: C
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 worksBTGmoderatorDC wrote: ↑Thu Apr 25, 2019 7:14 pmAt 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
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