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
Answer is D. Someone help with detailed explanation?
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Princeton -Probability
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tutonaranjo
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Total no of ways 5 people can sit = 5! = 120
Assume one couple as 1 entity. So, there are only 4 entities. This combination can sit in 4! ways.
2 person in the couple can sit in 2! ways.
So, total no of ways = 2!*4! = 48 ---------- (1)
Similarly for another set of couple, total no of ways = 48 ---------- (2)
Total no of ways both couple will sit together = 2*2*3! = 24 --------(3)
In (1) and (2) we have calculated (3) already. So we need to subtract it once from the total otherwise it will be counted twice.
So total no of ways (at least one couple sit together) = 48+48-24 = 72
So probability of at least one couple sitting together = 72/120
Probability (neither couple sits together) = 1-72/120 = 2/5
Assume one couple as 1 entity. So, there are only 4 entities. This combination can sit in 4! ways.
2 person in the couple can sit in 2! ways.
So, total no of ways = 2!*4! = 48 ---------- (1)
Similarly for another set of couple, total no of ways = 48 ---------- (2)
Total no of ways both couple will sit together = 2*2*3! = 24 --------(3)
In (1) and (2) we have calculated (3) already. So we need to subtract it once from the total otherwise it will be counted twice.
So total no of ways (at least one couple sit together) = 48+48-24 = 72
So probability of at least one couple sitting together = 72/120
Probability (neither couple sits together) = 1-72/120 = 2/5
