The thousands place can only be filled with 2,3 or 4, since the number is greater than 2000 . For the remaining 3 places, we have pick out digits such that the resultant number is divisible by 3 . It the sum of digits of the number is divisible by 3, then the number itself is divisible by 3 Case 1: If we take 2 at thousands place. The remaining digits can be filled as: 0,1 and 3 as 2+1+0+3=6 is divisible by 3 0,3 and 4 as 2+3+0+4=9 is divisible by 3 In both the above combinations the remaining three digits can be arranged in 3! ways. ∴ Total number of numbers in this case =2×3!=12. Case 2: If we take 3 at thousands place. The remaining digits can be filled as: 0,1 and 2 as 3+1+0+2=6 is divisible by 3 . 0,2 and 4 as 3+2+0+4=9 is divisible by 3 . In both the above combinations, the remaining three digits can be arranged in 3! ways. Total number of numbers in this case =2×3!=12 Case 3 : If we take 4 at thousands place. The remaining digits can be filled as: 0,2 and 3 as 4+2+0+3=9 is divisible by 3 . In the above combination, the remaining three digits can be arranged in 3! ways. ∴ Total number of numbers in this case =3!=6 ∴ Total number of numbers between 2000 and 5000 divisible by 3 are 12+12+6=30