Question

Six boys and six girls sit in a row at random. Find the probability that boys and girls sit alternately.

Answer 1

i) The number of ways in which all the 6 girls sit together is $6!\times 7!$ (considering all 6 girls as one person).Therefore,the probability of all girls sitting together is is (6! X7!)/12!-720/(12 x 11 x 10 x 9 x 8)=(1/132).

ii) Starting with a boy , boys can be sit in 6! ways leaving one place between every two boys and one at last

B_B_B_B_B_

These places can be occupied by girls in 6! ways.Therefore, if we start with a boy, number of ways of boys and girls sitting alternately is 6! x 6!

G_G_G_G_G_G_

Thus total number of ways of alternating sitting arrangements is 6!x 6!+6!x6!=2 x6! x6!

Therefore , the probability of making alternative sitting arrangement for 6 boys and 6 girls is

$\frac{2\times 6!\times 6!}{12!}=\frac{2times720}{12\times 11\times 10\times 9\times 8\times 7}=\frac{1}{462}$