problem of permutation

(4!*5!)/2!=1440

First we need to find the number of ways of arranging $4$ $b$ 's and $3$ $g$ 's so that no two $g$ 's are next to each other. Let $w$ be the number $b$ 's that appear prior to the first $g$ , $x$ the number of $b$ 's between the first and second $g$ , $y$ the number of $b$ 's between the second and third $g$ , and $z$ the number of $b$ 's after the third $g$ . Then we need to find the number of solutions to the equation

$w + x + y + z = 4$ where $w \ge 0, x \ge 1, y \ge 1, z \ge 0$ .

Now let $x' = x - 1$ and $y' = y - 1$ . We are then looking for the number of non-negative integer solutions to the equation

$w + x' + y' + z = 2$ ,

which is a 'stars and bars' calculation with solution $\binom{5}{2} = 10$ .

(This link describes the 'stars and bars' concept.)

Finally, for these $10$ solutions we can arrange the boys in $4!$ ways and the girls in $3!$ ways, giving us a final solution of $10 * 4! * 3! = \boxed{1440}$ .