GreenlightTestPrep wrote:
At a dinner party, 6 people (A, B, C, D, E, and F) are to be seated around the circular table shown above. Two seating arrangements are considered different only when the positions of the people are different relative to each other. If person B must sit directly across from person C (e.g., B in chair #1 and C in chair #4), what is the total number of different possible seating arrangements for the group?
(A) 20
(B) 24
(C) 48
(D) 72
(E) 144
Let’s solve this question via the most basic of counting techniques, the Fundamental Counting Principle (FCP):
As always, we’ll begin the most restrictive stage(s):
Stage 1: We can seat person B in
6 ways (seat B in chair 1, 2, 3, 4, 5 or 6).
Stage 2: At this point, we can seat person C in
1 way (directly across from person B).
Stage 3: We can seat person A in
4 ways (any remaining seat that isn’t yet occupied)
Stage 4: We can seat person D in
3 ways (any remaining seat that isn’t yet occupied)
Stage 5: We can seat person E in
2 ways (any remaining seat that isn’t yet occupied)
Stage 6: We can seat person F in
1 way
Now apply the Fundamental Counting Principle to see that the total number of ways to seat all 6 people = (
6)(
1)(
4)(
3)(
2)(
1) =
144Important: Since two seating arrangements are considered different only when the positions of the people are different relative to each other, we see that we’ve counted each unique outcome 6 times. For example, these 6 arrangements . . .
. . . are considered 1 unique arrangement because the relative positions of the six people are the same in each case. .
So, to account for counting each arrangement 6 times, we’ll divide
144 by 6 to get 24, which means the correct answer is B.
Answer: B
13 Apr 2022, 09:09
34% (01:53) correct
