All possible four-digit numbers have formed using the digits 0, 1, 2, 3 so that no number has repeated digits. The number of even numbers among them, is

Options

A. 9
B. 18
C. 10
D. none of these

Answer

To solve the problem of finding the number of four-digit even numbers that can be formed using the digits 0, 1, 2, and 3 without repeating any digits, we can follow these steps:

### Step 1: Identify the last digit
Since we want to form even numbers, the last digit must be either 0 or 2.

### Step 2: Case 1 - Last digit is 0
- If the last digit is 0, the first digit can be any of the remaining digits (1, 2, or 3).
- This gives us 3 choices for the first digit.
- After choosing the first digit, we have 2 remaining digits to choose from for the second position.
- Finally, we have 1 digit left for the third position.

So, the total number of combinations when the last digit is 0 is:
\[ \text{Choices} = 3 \text{ (for first digit)} \times 2 \text{ (for second digit)} \times 1 \text{ (for third digit)} = 3 \times 2 \times 1 = 6 \]

### Step 3: Case 2 - Last digit is 2
- If the last digit is 2, the first digit can be any of the remaining digits (1, 0, or 3).
- This gives us 2 choices for the first digit (it cannot be 0).
- After choosing the first digit, we have 2 remaining digits to choose from for the second position.
- Finally, we have 1 digit left for the third position.

So, the total number of combinations when the last digit is 2 is:
\[ \text{Choices} = 2 \text{ (for first digit)} \times 2 \text{ (for second digit)} \times 1 \text{ (for third digit)} = 2 \times 2 \times 1 = 4 \]

### Step 4: Total even numbers
Now, we can add the two cases together to find the total number of four-digit even numbers:
\[ \text{Total even numbers} = 6 \text{ (last digit 0)} + 4 \text{ (last digit 2)} = 10 \]

### Final Answer
The total number of four-digit even numbers that can be formed using the digits 0, 1, 2, and 3 without repeating any digits is **10**.

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