Question

Six boys and six girls sit in a row at random. Find the probability that
(i) the six girls sit together.
(ii) the boys and girls sit alternatively.

Answer 1

(i)The total number of arrangements of six boys and six girls $= 12! $
$\therefore $ Required probability =$\frac{6! \times 7!}{(12)!}=\frac{1}{132}$
[since, we consider six girls at one person]
(ii) Required probability =$\frac{2 \times 6! \times 6!}{(12)!}=\frac{1}{462}$