Step 1: Understanding the Concept:
Divisibility by 6 requires the number to be even and the sum of digits to be a multiple of 3.
Step 2: Key Formula or Approach:
Identify all possible 4-digit combinations that sum to a multiple of 3, then count permutations satisfying evenness and leading zero constraints.
Step 3: Detailed Explanation:
Digits: {0, 2, 3, 4, 6}. We choose 4 digits.
Cases for sum divisible by 3:
1. Digits {2, 3, 4, 6} (sum 15):
Ends in 2, 4, or 6 (3 choices). Permutations: \( 3 \cdot (3!) = 18 \).
2. Digits {0, 2, 4, 6} (sum 12):
End in 0: \( 3! = 6 \).
End in 2, 4, 6: \( 3 \text{ choices} \cdot (2 \cdot 2 \cdot 1) = 12 \).
Total for this set = 18.
3. Digits {0, 2, 3, 4} (sum 9):
End in 0: \( 3! = 6 \).
End in 2, 4: \( 2 \text{ choices} \cdot (2 \cdot 2 \cdot 1) = 8 \).
Total for this set = 14.
Overall Total \( = 18 + 18 + 14 = 50 \).
Step 4: Final Answer:
The number of such numbers is 50.