Digits available

We can form 4-digit numbers using the digits $\{0,1,2,3,4\}$ without repetition.

We want numbers:

• between $2000$ and $5000$
• divisible by $3$

Condition for being between 2000 and 5000

Since the number is 4-digit and must lie between $2000$ and $5000$, its thousands digit can only be: $2,3,4$

Condition for divisibility by 3

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

The available digits are $0,1,2,3,4$.

Their residues modulo $3$ are:

• $0 \equiv 0$
• $1 \equiv 1$
• $2 \equiv 2$
• $3 \equiv 0$
• $4 \equiv 1$

We must choose 4 distinct digits from these 5 digits.

Find which 4-digit selections give sum divisible by 3

Total sum of all 5 digits is: $0+1+2+3+4=10$ and $10 \equiv 1 \pmod{3}$

If we omit one digit $d$, then the sum of the chosen 4 digits is: $10-d$ For divisibility by $3$: $10-d \equiv 0 \pmod{3}$ $1-d \equiv 0 \pmod{3}$ $d \equiv 1 \pmod{3}$

Among $\{0,1,2,3,4\}$, the digits congruent to $1 \pmod{3}$ are: $1,4$

So the valid 4-digit sets are obtained by omitting either $1$ or $4$:

• Set 1: $\{0,2,3,4\}$
• Set 2: $\{0,1,2,3\}$

Count arrangements from each valid set with first digit $2,3,$ or $4$

Case 1: Digits $\{0,2,3,4\}$

Total permutations of these 4 digits: $4! = 24$

But numbers cannot start with $0$. Number starting with $0$: $3! = 6$

Hence valid 4-digit numbers from this set: $24-6=18$

All these have first digit among $2,3,4$, so all lie between $2000$ and $5000$.

Case 2: Digits $\{0,1,2,3\}$

Total permutations: $4! = 24$

Numbers starting with $0$: $3! = 6$

So 4-digit numbers formed: $24-6=18$

But to be between $2000$ and $5000$, first digit must be $2$ or $3$ only.

Count directly:

• First digit $2$: remaining 3 digits can be arranged in $3! = 6$ ways
• First digit $3$: remaining 3 digits can be arranged in $3! = 6$ ways

Total for this case: $6+6=12$

Total count

$18+12=30$

Compare with options

The required number is: $\boxed{30}$

So the correct option is B.