Each of the digits 2, 3, 5 and 7 is used exactly once to build a four-digit number, and every such number is equally likely. Find the probability that the number built is a multiple of 3.

A number is divisible by 3 exactly when its digit sum is, and rearranging digits never changes their sum. Here the sum is 2 + 3 + 5 + 7 = 17 for every one of the 4! = 24 arrangements, and 17 is not a multiple of 3, so not a single arrangement works and the probability is 0. This is an impossible event, and spotting it takes one addition rather than any listing. The value 1/3 assumes that one number in every three must be a multiple of 3, which is true of a random stretch of integers but not of a set of numbers all sharing the same digit sum. The value 1/4 would be the answer to the divisibility-by-5 version, since only the 3! = 6 arrangements ending in 5 qualify. The value 1/24 counts a single arrangement out of the 24.