At a dinner party, 6 people (A, B, C, D, E, and F) are to be seated ar

Question

https://i.imgur.com/l5p3OZH.png

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

(A) 20
(B) 24
(C) 48
(D) 72
(E) 144

vaibhav1221 · Accepted Answer

The total number of ways in which n people can be arranged in a circular arrangement is (n-1)!
So, 6 people can be arranged in 5! ways or in 120 ways.
Now, assume that B sits in (1), C now has 5 options - (2), (3), (4), (5), (6) - out of which only (3) works.
This means that the condition would be true in 1/5th of the total cases.
So, total cases = 5!/5 = 4! = 24

MissBong · Answer

B  has to sit opposite  C , so we can fix their position.

Now, the rest i.e. A, D, E & F can take any of the 4 positions left.

Hence, the question becomes: In how many ways can four people sit at four places? =  4!  ways.

Therefore, the answer is  4! = 24 .

Option (B).

BrentGMATPrepNow · Answer

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  ) =   144

Important: 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 . . . 
 https://i.imgur.com/ttVfCKe.png 
. . . 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

RELATED VIDEO
Fundamental Counting Principle (FCP)
 https://www.youtube.com/watch?v=w5d5bvCRyd4

Fundamental Counting Principle - example
 https://www.youtube.com/watch?v=_PFBp2BoMUY

BrentGMATPrepNow · Answer

Great reasoning, Vaibhav!! (super short solution too!)
Top drawer mathing!

bumpbot · Answer

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At a dinner party, 6 people (A, B, C, D, E, and F) are to be seated ar

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At a dinner party, 6 people (A, B, C, D, E, and F) are to be seated ar

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