The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is:

We need to find how many 4-digit numbers between 2000 and 5000, formed using digits $$\{0, 1, 2, 3, 4\}$$ without repetition, are divisible by 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of all five available digits is $$0 + 1 + 2 + 3 + 4 = 10$$.

Since we pick 4 digits out of 5, we leave out one digit. The sum of the chosen 4 digits is $$10 - d$$, where $$d$$ is the digit left out.

For divisibility by 3: $$10 - d \equiv 0 \pmod{3}$$, which gives $$d \equiv 1 \pmod{3}$$.

From $$\{0, 1, 2, 3, 4\}$$, the digits with $$d \equiv 1 \pmod{3}$$ are $$d = 1$$ and $$d = 4$$.

Case 1: Leave out digit 1. Available digits: $$\{0, 2, 3, 4\}$$. Sum = 9, divisible by 3.

The first digit (thousands place) must be 2, 3, or 4 (to keep the number between 2000 and 4999). So the first digit has 3 choices.

The remaining 3 positions can be filled by the remaining 3 digits in $$3! = 6$$ ways.

Numbers in this case: $$3 \times 6 = 18$$.

Case 2: Leave out digit 4. Available digits: $$\{0, 1, 2, 3\}$$. Sum = 6, divisible by 3.

The first digit must be 2 or 3 (cannot be 0, and 4 is not available). So the first digit has 2 choices.

The remaining 3 positions can be filled by the remaining 3 digits in $$3! = 6$$ ways.

Numbers in this case: $$2 \times 6 = 12$$.

Total count = $$18 + 12 = 30$$.

The answer is Option B: 30.