What is the sum of all four digit numbers formed by using all digits 0, 1, 4, 5 without repetition of digits?

Total 4-digit numbers using 0,1,4,5 without repetition with non-zero leading digit: 3 × 3! = 18 numbers. Sum of digits = 0+1+4+5 = 10. Each non-zero digit (1,4,5) appears in thousands place 3! = 6 times. Digit 0 appears 0 times in thousands. For other positions (hundreds, tens, units): each digit appears (total-numbers-with-this-digit-fixed) times. Using standard formula: sum = (sum of digits)(n-1)! × (1111) − (correction for zero in leading). Sum = (1+4+5)×6×1000 + (0+1+4+5)×(arrangements where 0 not leading)×(100+10+1). Each of 1,4,5 in thousands 6 times: contribution 10×6×1000 = 60000. For hundreds: each digit appears 4 times (since one of three remaining slots, with leading non-zero choices). Computed sum equals 46440.