Explanation

(b)
First we fix the position of a girl between one place. Girls can be arranged in 3! ways.
There are four alternate places to be left. Therefore four boys occupy the place in 4! ways.
$\therefore$ The favourable cases $= 4! \times 3!$
Hence, the required probability
$= \frac{4! \times 3!}{7!} = \frac{6}{7 \times 6 \times 5} = \frac{1}{35}$