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Step-by-Step Solution
Step 1: Understand the Problem
We want four-digit numbers between 2000 and 5000 that use the digits 0, 1, 2, 3, 4 without repetition. Additionally, the four-digit number must be divisible by 3. We need to count how many such numbers exist.
Step 2: Recall the Divisibility Rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3. Hence, we look for sets of four distinct digits (from 0, 1, 2, 3, 4) whose sum is a multiple of 3.
Step 3: Identify Possible Digit Sets
We consider all 4-digit combinations from the digits 0, 1, 2, 3, 4 and find their sums:
Set (0, 2, 3, 4): sum = 0 + 2 + 3 + 4 = 9 (which is divisible by 3)
Set (0, 1, 2, 3): sum = 0 + 1 + 2 + 3 = 6 (which is divisible by 3)
These two sets of digits satisfy the divisibility-by-3 condition.
Step 4: Count Valid 4-Digit Numbers for Each Set
Case 1: Digits (0, 2, 3, 4)
The first digit must be 2, 3, or 4 to ensure the number is between 2000 and 5000. We must not use 0 as the first digit. The total number of ways to form the 4-digit number can be counted as shown in the original solution. Preserving the given image and combination notation:
The number of permutations is calculated as:
$$
{^3C_1} \times {^3C_1} \times {^2C_1} \times {^1C_1} = 3 \times 3 \times 2 \times 1 = 18.
$$
Case 2: Digits (0, 1, 2, 3)
Here the first digit can be 2 or 3 to ensure the number is between 2000 and 5000 (we cannot start with 0 or 1). Again, preserving the original image and notation:
The number of permutations is calculated as:
$$
{^2C_1} \times {^3C_1} \times {^2C_1} \times {^1C_1} = 2 \times 3 \times 2 \times 1 = 12.
$$
Step 5: Sum Up the Total Valid Numbers
Combining both cases gives:
$$
\text{Total} = 18 + 12 = 30.
$$
So, there are 30 such 4-digit numbers between 2000 and 5000 that can be formed using the given digits without repetition and are multiples of 3.
