Question

The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not allowed, is _________.

Hint:

Always handle '0' as a special case in permutation problems involving digit placement, as it cannot be the leading digit of a multi-digit number.

Answer 1

Correct Answer: 52

Step 1: Understanding the Concept:
To form a three-digit even number, the unit's digit must be even. In our set $\{0, 1, 3, 4, 6, 7\}$, the even digits are 0, 4, and 6. Since repetition is not allowed and the hundred's digit cannot be 0, we must consider the cases where the unit's digit is 0 separately.
Step 2: Detailed Explanation:
Case 1: Unit's digit is 0.
- The unit's place is fixed with '0' (1 way).
- For the hundred's place, any of the remaining 5 digits $\{1, 3, 4, 6, 7\}$ can be chosen (5 ways).
- For the ten's place, any of the remaining 4 digits can be chosen (4 ways).
- Total numbers $= 1 \times 5 \times 4 = 20$.
Case 2: Unit's digit is 4.
- The unit's place is fixed with '4' (1 way).
- For the hundred's place, the digit cannot be 0 or 4. The available digits are $\{1, 3, 6, 7\}$ (4 ways).
- For the ten's place, we can now include 0 but excluding the two digits already used. Remaining digits are $6 - 2 = 4$ ways.
- Total numbers $= 1 \times 4 \times 4 = 16$.
Case 3: Unit's digit is 6.
- This case is identical to Case 2.
- Total numbers $= 16$.
Final Calculation:
Total three-digit even numbers $= 20 + 16 + 16 = 52$.
Step 3: Final Answer:
The number of three-digit even numbers is 52.