Bunuel wrote:
9 people, including 3 couples, are to be seated in a row of 9 chairs. What is the probability that only one couple is sitting together
A. 1/63
B. 1/6
C. 8/21
D. 61/126
E. 7/9
Similar questions:
https://gmatclub.com/forum/9-people-inc ... 80326.htmlhttps://gmatclub.com/forum/9-people-inc ... 80328.htmlWe have 3 singles and 3 couples in all.
Total number of ways of arranging all 9 people = 9!
Use the Venn diagram of 3 overlapping sets to understand this.
All three couples, say couples A, B and C, sit together (overlap of all 3):
Make each couple a pair and arrange 6 people/pairs in 6! ways. Each couple can sit in 2 ways so 6!*2*2*2 ways = 6! * 8 ways.
Attachment:
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2 given couples (say couples A and B) sit together (entire overlap of A and B):
Make the two couples 2 pairs and arrange 2 pairs and 5 others in 7! ways. Each couple can sit in 2 ways so we get 7! * 2 * 2 ways.
This includes the number of ways in which all 3 couples are sitting together too.
So ONLY A and B sit together in 7! *2 * 2 - 6!*2*2*2 = 20*6! ways
The same will be applicable to B&C and C&A too.
Attachment:
Screenshot 2022-01-31 at 12.35.45.png [ 53.01 KiB | Viewed 1867 times ]
Couple A sits together (the whole yellow circle):
Arrange 1 pair and 7 others in 8! ways. Couple A can sit in 2 ways so we get 8! * 2 ways.
This includes the ways in which B and C sit together too (the overlap regions).
So only couple A sits together in 8! * 2 - (20*6!) - (6! * 8) - (20 * 6!) = 64 * 6! (only the region that is yellow)
Exactly one couple sits together (couple A or B or C) = 64 * 6! * 3 ways (yellow, blue & red regions)
Required Probability = \(\frac{(64 * 6! * 3) }{ 9!} = \frac{8}{21}\)
Answer (C)