Three fair dice, one red, one blue and one green, are thrown together. What is the probability that the red, blue and green numbers are in strictly increasing order, that is red < blue < green?

There are $6^3=216$ equally likely outcomes. For a strictly increasing reading the three numbers must all differ, and once three different numbers have been chosen there is exactly ONE way to place them in increasing order. So the favourable outcomes number $C(6,3)=20$, and the probability is 20/216 = 5/54. A second route: the three numbers are all different with probability $(6\times 5\times 4)/216=5/9$, and given that, only 1 of their 3! = 6 orders increases, so $(5/9)(1/6)=5/54$. The value 5/9 stops at 'all three different'. The value 1/6 applies the ordering step but forgets that ties must first be excluded — with repeats allowed there is no meaningful ordering to count. The value 7/27 = 56/216 counts non-DECREASING readings, allowing ties, from $C(8,3)=56$.