Question

Six boys and six girls sit in a row randomly. The probability the boys and the girls sit alternatively is

Answer 1

Starting with a boy, boys can sit in $6!$ ways leaving one place between every two boys and one at the last.
B_B_B_B_B_
These places can be occupied by girls in $6!$ ways. Therefore, if we start with a boy, the number of ways of boys and girls sitting alternatively is $6!\times 6!$.
G_G_G_G_G_G_
Thus, total number of ways of alternative sitting arrangements is $6!\times 6!+6!\times 6!=2\times 6!\times 6!$
Thus, the probability of making alternative sitting arrangements for $6$ boys and $6$ girls is
$\frac{2\times 6!\times 6!}{12!}=\frac{2\times 720}{12\times 11\times 10\times 9\times 8\times 7}=\frac{1}{462}$