Six boys and six girls sit in a row randomly. Find the probability that (i) the six girls sit together, (ii) the boys and girls sit alternately.

Explanation

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!×6!.

G_G_G_G_G_G_

Thus, total number of ways of alternative sitting
arrangements is `(6!×6!+6!×6!)=(2×6!×6!)`

Thus, the probability of making alternative sitting
arrangements for 6 boys and 6 girls is

`(2xx6!xx6!)/(12!)`

`=(2xx720)/(12xx11xx10xx9xx8xx7)`

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