Key Concepts and Formulas

• Divisibility Rule for 3: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
• Permutations: The number of ways to arrange $n$ distinct objects in a specific order is given by $n!$. The number of ways to choose and arrange $k$ objects from $n$ distinct objects is $P(n, k) = \frac{n!}{(n-k)!}$.
• Problem Constraints:
  • The number must be a 4-digit number between 2000 and 5000.
  • The first digit must be 2, 3, or 4.
  • Digits available are 0, 1, 2, 3, and 4.
  • Repetition of digits is not allowed.

Step-by-Step Solution

Let the 4-digit number be represented as $d_1 d_2 d_3 d_4$, where $d_1$ is the thousands digit, $d_2$ is the hundreds digit, $d_3$ is the tens digit, and $d_4$ is the units digit.

Step 1: Determine the sets of 4 digits that satisfy the divisibility rule for 3.

We are given the digits $S = \{0, 1, 2, 3, 4\}$. We need to choose 4 distinct digits from this set such that their sum is a multiple of 3. The sum of all available digits is $0 + 1 + 2 + 3 + 4 = 10$. If we select 4 digits, it means we are excluding exactly one digit from the set $S$. Let the excluded digit be $x$. The sum of the four chosen digits will be $10 - x$. We need $10 - x$ to be divisible by 3.

Let's test each possible value for $x$:

• If $x = 0$, sum $= 10 - 0 = 10$ (not divisible by 3).
• If $x = 1$, sum $= 10 - 1 = 9$ (divisible by 3). This means the set of digits is $\{0, 2, 3, 4\}$.
• If $x = 2$, sum $= 10 - 2 = 8$ (not divisible by 3).
• If $x = 3$, sum $= 10 - 3 = 7$ (not divisible by 3).
• If $x = 4$, sum $= 10 - 4 = 6$ (divisible by 3). This means the set of digits is $\{0, 1, 2, 3\}$.

So, there are two possible sets of 4 digits that can form numbers divisible by 3:

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

Step 2: Form 4-digit numbers using Set A, subject to all constraints.

The digits available are $\{0, 2, 3, 4\}$. The number must be between 2000 and 5000. This means the thousands digit ($d_1$) can only be 2, 3, or 4.

Let's place the digits:

• For $d_1$ (thousands digit): From Set A, the possible choices for $d_1$ are 2, 3, or 4. (3 choices)
  • Why? Because $d_1$ cannot be 0 (otherwise it's not a 4-digit number), and it must be $\ge 2$ and $< 5$ to be within the range [2000, 5000). All digits 2, 3, 4 are available in Set A.
• For $d_2$ (hundreds digit): After choosing $d_1$, 3 digits remain from Set A. (3 choices)
  • Why? Repetition is not allowed, so one digit is used for $d_1$. The remaining 3 digits can be placed in $d_2$.
• For $d_3$ (tens digit): After choosing $d_1$ and $d_2$, 2 digits remain from Set A. (2 choices)
  • Why? Two digits are already used, so 2 distinct digits are left.
• For $d_4$ (units digit): After choosing $d_1, d_2, d_3$, 1 digit remains from Set A. (1 choice)
  • Why? Three digits are already used, leaving only 1.

Total numbers formed using Set A = $3 \times 3 \times 2 \times 1 = 18$.

Step 3: Form 4-digit numbers using Set B, subject to all constraints.

The digits available are $\{0, 1, 2, 3\}$. The number must be between 2000 and 5000. This means the thousands digit ($d_1$) can only be 2, 3, or 4.

Let's place the digits:

• For $d_1$ (thousands digit): From Set B, the possible choices for $d_1$ are 2 or 3. (2 choices)
  • Why? $d_1$ must be from $\{2, 3, 4\}$. However, the digit 4 is NOT present in Set B. Thus, $d_1$ can only be 2 or 3.
• For $d_2$ (hundreds digit): After choosing $d_1$, 3 digits remain from Set B. (3 choices)
  • Why? One digit is used for $d_1$. The remaining 3 digits can be placed in $d_2$.
• For $d_3$ (tens digit): After choosing $d_1$ and $d_2$, 2 digits remain from Set B. (2 choices)
  • Why? Two digits are already used, so 2 distinct digits are left.
• For $d_4$ (units digit): After choosing $d_1, d_2, d_3$, 1 digit remains from Set B. (1 choice)
  • Why? Three digits are already used, leaving only 1.

Total numbers formed using Set B = $2 \times 3 \times 2 \times 1 = 12$.

Step 4: Calculate the total number of such numbers.

The total number of numbers satisfying all conditions is the sum of numbers from Set A and Set B. Total numbers = (Numbers from Set A) + (Numbers from Set B) Total numbers = $18 + 12 = 30$.

Common Mistakes & Tips

• Missing Constraints: Always explicitly state all constraints before starting the solution. Overlooking even one constraint can lead to an incorrect answer.
• Leading Zero: Remember that the first digit of a multi-digit number cannot be zero. This is implicitly taken care of by the range constraint in this problem.
• Systematic Approach: Break the problem into manageable cases. This reduces the chance of errors and ensures all possibilities are considered. Pay close attention to the first digit's restrictions.

Summary

We identified the possible sets of 4 digits from $\{0, 1, 2, 3, 4\}$ whose sum is divisible by 3: $\{0, 2, 3, 4\}$ and $\{0, 1, 2, 3\}$. Then, for each set, we constructed 4-digit numbers, ensuring the first digit was between 2 and 4 (inclusive) and that no digit was repeated. By summing the results from these two cases, we obtained the final count.

The final answer is $\boxed{30}$, which corresponds to option (B).