mohan514 wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?
A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2
Responding to a pm:
I am guessing you are looking for a one line solution since the answer is a simple 2/5. I couldn't think of the logic that will help us arrive at the answer directly. I use sets to solve such questions. It's not very different from what Bunuel has used above.
Here is my solution:
This question isn't very different from our regular "5 people sit in a row such that A does not sit next to B. In how many ways is this possible?"
Here, there are 2 pairs of people who cannot sit next to each other. Hence, you need to take special care of the cases in which both sit with each other.
There are two couples. We don’t want either couple to sit together. Let’s go the reverse way – let’s make at least one of them sit together. We can then subtract this number from the total arrangements to get the number of arrangements in which neither couple sits together.
Would you agree that it is easy to find the number of arrangements in which both couples sit together? It is. We will work on it in a minute. Let’s think ahead for now.
How about ‘finding the number of ways in which one couple sits together?’ Sure we can easily find it but it will include those cases in which both couples are sitting together too. But we would have already found the number of ways in which both couples sit together. When we just subtract ‘both couples together’ number once from the total to avoid double counting, we will get the number of ways in which at least one couple sits together. Think of SETS here.
Let’s do this now.
Number of arrangements in which both couples sit together: Let’s say the two couples are {C1h, C1w} and {C2h, C2w} and the single person is S. There are three groups/individuals. They can be arranged in 3! ways. But in each couple, husband and wife can be arranged in 2 ways (husband and wife can switch places)
Hence, number of arrangements such that both couples are together = 3!*2*2 = 24
Number of arrangements such that C1h and C1w are together: C1 acts as one group. We can arrange 4 people/groups in 4! ways. C1h and C1w can be arranged in 2 ways (husband and wife can switch places).
Number of arrangements in which C1h and C1w are together = 4! * 2 = 48
But this 48 includes the number of arrangements in which C2h and C2w are also sitting together.
C2h and C2w also sit together in 48 ways (including the number of ways in which C1h and C1w also sit together)
Number of arrangements in which at least one couple sits together = 48 + 48 - 24 = 72
Number of arrangements in which neither couple sits together = 120 – 72 = 48
Probability that neither couple sits together = 48/120 = 2/5
There is an alternative solution of case by case evaluation discussed here:
http://www.veritasprep.com/blog/2012/01 ... e-couples/Thanks Karishma for taking the efforts to explain this. +1