Question

From the digits 0, 1, 2, 3, three-digit numbers are formed. Among that, the number of even numbers are:

• A. 6
• B. 10
• C. 12
• D. 18

Hint:

Always check for leading digit constraints and consider each unit digit case separately.

Answer 1

The Correct Option is B

We are asked to form 3-digit numbers using the digits: 0, 1, 2, 3 (no repetition). For a number to be even, the unit digit must be even — i.e., either 0 or 2. Let's consider both cases: Case 1: Unit digit is 0
Then we need to choose 2 digits from \{1, 2, 3\} for the hundreds and tens place.
Number of ways to choose and arrange = \( 3 \times 2 = 6 \) Case 2: Unit digit is 2
Now the remaining digits are \{0, 1, 3\}. Hundreds digit cannot be 0.
Let's count valid combinations:

Hundreds digit = 1, tens = 0/3 → 2 options
Hundreds digit = 3, tens = 0/1 → 2 options
Hundreds digit = 0 → not valid (leading zero not allowed)

So, number of valid numbers = 2 + 2 = 4 Total even numbers = 6 (when unit is 0) + 4 (when unit is 2) = \fbox{10}