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?

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[tex]N=8!.[/tex]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[tex]n=4!\times4!\times2!.[/tex]Therefore, the probability that all the women will sit in adjacent seats and all the men will sit in adjacent​ seats is[tex]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\%.[/tex]Thus, the required probability is 2.86%.