The sum of all numbers greater than 1000 formed by using the digits 0, 1, 2, 3, no digit being repeated in any number, is

detailed solution

Correct option is A

All the numbers are of four digits and they do not have 0 in thousand’s place.The number of numbers having 0 in unit’s place = 3! (∵ the other three places are to be filled by 1, 2 or 3).The number of numbers having 1 in unit’s place =2P1×2P2(∵ thousand’s place can be filled by one of 2, 3 and the remaining two places can be filled by the remaining two digits).Similarly, the number of numbers having 2 or 3 in unit’s place=2P1×2P2 in each case.Thus, the sum of the digits in unit’s place for all the numbers=3!×0+2P1×2P2×1+2P1×2P2×2+2P1×2P2×3=4+8+12=24Similarly, the sum of the digits in ten’s and hundred’s places = 24 each.Now, the thousand’s place can have only 1 or 2 or 3.Number of numbers having 1 in thousand’s place = 3!, (for the other three places will be filled by 0, 2, 3).Similarly, the number of numbers having 2 or 3 in thousands’s place is 3! in each case.∴ Sum of the digits in the thousands’s place for all the numbers = 3! × 1 + 3! × 2 + 3! × 3 = 6 + 12 + 18 = 36.∴ Required sum of all the numbers= 36 × 1000 + 24 × 100 + 24 × 10 + 24 × 1= 36000 + 2400 + 240 + 24 = 38664.