Five boys and four girls are to be seated in a row so that two particular girls will never sit adjacent to a particular boy and all girls are separated. If the number of ways in which they can be seated is N, the value of N/(480) is?

See below:

Explanation:

If I'm reading this right, we have 5 boys and 4 girls and are seated in a row such that the girls are all separated:

$B \textcolor{w h i t e}{0} G \textcolor{w h i t e}{0} B \textcolor{w h i t e}{0} G \textcolor{w h i t e}{0} B \textcolor{w h i t e}{0} G \textcolor{w h i t e}{0} B \textcolor{w h i t e}{0} G \textcolor{w h i t e}{0} B$

With no other restrictions, the number of ways we can arrange this grouping is:

N=5!xx4! = 120xx24=2880

At this point, $\frac{N}{480} = 6$

For the language about 2 particular girls not sitting next to a particular boy, we can put that restriction in. Let's first freeze the two girls (they'll be the first two Gs). We don't want the boy to be the second B:

$B \textcolor{w h i t e}{0} \textcolor{\mathmr{and} a n \ge}{G} \textcolor{w h i t e}{0} \textcolor{red}{B} \textcolor{w h i t e}{0} \textcolor{\mathmr{and} a n \ge}{G} \textcolor{w h i t e}{0} B \textcolor{w h i t e}{0} G \textcolor{w h i t e}{0} B \textcolor{w h i t e}{0} G \textcolor{w h i t e}{0} B$

There are 3 positions along the row that the two girls can sit.

There are 2 ways to arrange the two girls in those seats.

There are 2 ways that the other two girls can sit in the remaining 2 seats. This gives the number of ways to arrange the girls is:

$3 \times 2 \times 2 = 12$

For the boys, the first boy can only sit in one of 4 seats. The remaining boys can sit in any of the remaining 4 seats, giving:

4xx4! = 4xx24=96

And so we get:

$N = 12 \times 96 = 1152 \implies \frac{N}{480} = 2.4$