In how many ways can 5 boys and 3 girls be arranged in a row so that no two girls stand next to each other?

First seat the 5 boys in 5! = 120 ways. That creates 6 gaps (including the two ends); place the 3 girls in distinct gaps in $6\times 5\times$ 4 = 120 ways. Total = $120\times 120=14400$. The value 2880 = 5! $x$ 4! is the count when the girls are kept together instead.