Question

Four married couples have reserved eight seats in a row at the theater. If they arrange themselves​ randomly, what is the probability that all the women will sit in adjacent seats and all the men will sit in adjacent​ seats?

Answer 1

The required probability is 2.86%.

Step-by-step explanation:  

The number of ways in which eight persons can sit within themselves at the reserved eight seats is given by

$N=8!.$

And, the number of ways in which 4 couples (8 persons) can sit such that all the women will sit in adjacent seats and all the men will sit in adjacent​ seats is given by

$n=4!\times4!\times2!.$

Therefore, the probability that all the women will sit in adjacent seats and all the men will sit in adjacent​ seats is

$p=\dfrac{n}{N}=\dfrac{4!\times4!\times2!}{8!}=\dfrac{4\times3\times2\times1\times4!\times2\times1}{8\times7\times6\times5\times4!}=\dfrac{1}{35}=2.86\%.$

Thus, the required probability is 2.86%.