Six-digit numbers are formed using the digits 1, 2, 3, 4, 5, and 6 without repetition. If all these numbers are arranged in strictly increasing order, then the number located at the 425^ th position is :

Step-by-step solution

The digits are 1, 2, 3, 4, 5, 6 . The number of 6-digit numbers starting with 1 is 5! = 120 . The number of 6-digit numbers starting with 2 is 5! = 120 . The number of 6-digit numbers starting with 3 is 5! = 120 . Total numbers formed so far = 120 + 120 + 120 = 360 . The target is the 425^ th number, so it must start with 4 . The number of numbers starting with 41 is 4! = 24 . The number of numbers starting with 42 is 4! = 24 . Total numbers formed so far = 360 + 24 + 24 = 408 . The number of numbers starting with