The number of numbers between 2000 and 5000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiples of 3 is?

There are 4 places to be filled with the given digits. The thousands place can have only 2, 3 and 4 since the number has to be greater than 2000. For the remaining 3 places, we have to pick out digits such that the resultant number is divisible by 3. The divisibility criteria for 3 states that the sum of digits of the number should be divisible by 3.
Case 1: If we pick 2 for the thousands place. The remaining digits we can pick such that the sum of digits at all places is a multiple of 3 are: 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 = 2 × 3! = 12
Case 2: If we pick 3 for the thousands place. The remaining digits we can pick such that the sum of digits at all places is a multiple of 3 are: 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 = 2 × 3! = 12
Case 3: If we pick 4 for the thousands place. The remaining digits we can pick such that the sum of digits at all places is a multiple of 3 are: 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 = 3! = 6
Total number of numbers between 2000 and 5000 divisible by 3 are 12 + 12 + 6 = 30. Option A is correct.