Six-digit numbers are formed using the digits 0, 1, 3, 5, 7, 9 without repetition. If all such valid numbers are arranged in increasing order of their magnitude, then the 400^ th number in this arrangement is :

Step-by-step solution

The given digits are 0, 1, 3, 5, 7, 9 . We are forming 6-digit numbers, so 0 cannot be the first digit. Number of valid 6-digit numbers starting with 1 = 5! = 120 Number of valid 6-digit numbers starting with 3 = 5! = 120 Number of valid 6-digit numbers starting with 5 = 5! = 120 Total numbers so far = 120 + 120 + 120 = 360 The 400^ th number must start with 7 . Number of numbers starting with 70 = 4! = 24 (Total = 384 ) Number of numbers starting with 71 = 4! = 24 (Total = 408 ) So, the 400^ th number starts with