The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is

The total number of numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time =4P4=4!=24. Consider the digit in the unit’s place in all these numbers. Each of the digits 2, 3, 4, 5 occurs in 3! = 6 times in the unit’s place∴ total for the digits in the unit’s place = (2 + 3 + 4 + 5) 6 = 84Since each of the digits 2, 3, 4, 5 occurs 6 times in any one of the remaining places∴ the required total = 84 (1 + 10 + 102 + 103) = 84 (1111) = 93324.