Three boys and three girls stand in a row in a random order. What is the probability that no two boys stand next to each other?

All six children can stand in 6! = 720 orders. Place the 3 girls first (3! = 6 ways); they create 4 gaps _G_G_G_ and the boys must occupy 3 different gaps, which can be done in $C(4,3)\times 3$! = $4\times 6=24$ ways. Favourable orders = $6\times 24=144$, so the probability is 144/720 = 1/5.