Question

You need to follow the comments

Three couple sits randomly in a row. What is the probability that no husband sits beside his wife? (you need to use inclusion-exclusion formula and let all the event, such as P(AUB), P(A intersection B formula to calculate it, hints P(A)=1-P(A^c) )

Answer 1

16:42 There are three ways to select a couple who sit in adjacent seats. That gives us five objects to arrange, the couple and the other four people The objects can be arranged in 5! ways. The couple that sits together can be arranged internally in 2! ways. Hence, there are 512! seating arrangements in which a couple sits in adjacent seats However, if we subtract these seating arrangements from the total, we will have subtracted too much since we have counted seating arrangements in which two couples sit together twice, once for each way we could designate one of the couples as the couple that sits in adjacent seats. Since we only want to subtract such couples once, we must add them back There are (3) ways to select two couples th at sit together. That gives us four objects to arrange, the two couples and the two other people. The objects can be arranged in 4! ways. Each of the two couples that sit in adiacent seats can be arranged internally in 2! ways. Hence, there are 4!2!2! seating arrangements in which two couples sit together When we subtracted arrangements in which a couple sits together, we counted seating arrangements in which all three couples sit

16:43 When we subtracted arrangements in which a couple sits together, we counted seating arrangements in which all three couples sit together three times, once for each way we could have designated one of those couples as the couple that sits together. When we added arrangements in which two couples sit together, we counted seating arrangements in which all three couples sit together three times, once for each of the () ways we could have designated two of the three couples as the ones that sit together. Therefore, we have not excluded seating arrangements in which all three couples sit together at all There are 3! ways to arrange three couples Each couple can be arranged internally in 2! ways. Hence, the number of seating arrangements in which all three couples sit together is 3!212!2! By the Inclusion-Exclusion Principle, the number of seating arrangements of the three couples in which no couples sit together is 3!2!2!2! The probability that no couple sits together is 6!

16:43 times, once ror eacn or tne (2) Ways we couia have designated two of the three couples as the ones that sit together. Therefore, we have not excluded seating arrangements in which all three couples sit together at all There are 3! ways to arrange three couples Each couple can be arranged internally in 2! ways. Hence, the number of seating arrangements in which all three couples sit together is 3!21212! By the Inclusion-Exclusion Principle, the number of seating arrangements of the three couples in which no couples sit together is 4!2!2! 3!212!2! The probability that no couple sits together is 512!+ 3!2!2!2! 6! 5!2! 21232122! 6!

Hence required probability = 1/3