Question

The number of 3-digit even numbers that can be formed with the digits 0, 1, 2, 3, 4, 5, 6 are

• A. 30
• B. 75
• C. 105
• D. 140
• E. 150

Hint:

Logic Tip: Whenever zero is included in a set of digits used to form even numbers (or numbers divisible by 5), ALWAYS split the calculation into two cases: one where zero is the last digit, and one where it isn't. This prevents overcounting or undercounting the restrictions on the first digit.

Answer 1

The Correct Option is C

Concept:
This is a permutations problem where repetition of digits is not allowed. A number is a valid 3-digit number if its hundreds digit is not 0. A number is even if its units digit is 0, 2, 4, or 6. Because the digit '0' is restricted in two different places, we must separate the problem into two distinct cases.
Step 1: Case 1: The number ends in 0.
Units place: 1 choice (the digit 0). Hundreds place: Since 0 is already used, we have 6 remaining non-zero digits (1, 2, 3, 4, 5, 6), so there are 6 choices. Tens place: 5 remaining digits to choose from. $$\text{Combinations for Case 1} = 6 \cdot 5 \cdot 1 = 30$$
Step 2: Case 2: The number ends in 2, 4, or 6.
Units place: 3 choices (2, 4, or 6). Hundreds place: We cannot use 0, and we cannot use the digit already placed in the units place. Out of 7 total digits, 2 are restricted, leaving 5 choices. Tens place: We cannot use the units digit or the hundreds digit, but 0 is now allowed again. Out of 7 total digits, 2 are used, leaving 5 choices. $$\text{Combinations for Case 2} = 5 \cdot 5 \cdot 3 = 75$$
Step 3: Calculate the total combinations.
Add the combinations from both mutually exclusive cases: $$\text{Total} = 30 + 75 = 105$$