Three 10-sided dice are rolled. If each die has a different digit from 0 to 9, inclusive, on each face, what is the probability that the three digits rolled can be arranged to create one of the possible 3-digit values of N from the previous problem? Express your answer as a common fraction.
There are 10 × 10 × 10 = 1000 possible rolls. Rolls from 000 to 999 are possible. There are 14 values for N, as we found in the previous problem, but since the digits rolled can be rearranged, there is more than one way to achieve each of these values. Of these 14 values, 11 have three different digits (ex: 126), however, there are two pairs of numbers that contain the same three digits, 189 and 819, as well as, 567 and 756. This means there are 11 – 2 = 9 unique combinations of three distinct digits. There are 3! = 6 ways to arrange the three numbers, therefore, there are 6 × 9 = 54 ways to roll and obtain one of these numbers. Of these 14 values, 3 have two of the same digit (ex: 441). For the numbers with two of the same digit there are 3! ÷ 2! = 3 ways to arrange the numbers, and 3 × 3 = 9 ways to roll and obtain on of these numbers. The probability of rolling three numbers that can be arranged to form on of the values of N is (54 + 9)/1000 = 63/1000.
