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

Answer

To solve the problem of counting the number of four-digit even numbers that can be formed using the digits 0, 1, 2, and 3 without repetition, we will break it down into two cases based on the last digit (which must be even).

### Step-by-Step Solution:

1. **Identify the even digits**: The even digits available from the set {0, 1, 2, 3} are 0 and 2.

2. **Case 1: Last digit is 0**:
- If the last digit is 0, the first digit can only be 1, 2, or 3 (it cannot be 0 because it is a four-digit number).
- Therefore, we have 3 choices for the first digit (1, 2, or 3).
- 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.
- The total number of arrangements for this case is:
\[ \text{Choices for first digit} \times \text{Choices for second digit} \times \text{Choices for third digit} = 3 \times 2 \times 1 = 6 \]

3. **Case 2: Last digit is 2**:
- If the last digit is 2, the first digit can be 1, 3, or 0 (it cannot be 2).
- Therefore, we have 2 choices for the first digit (1 or 3, since 0 cannot be the leading 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.
- The total number of arrangements for this case is:
\[ \text{Choices for first digit} \times \text{Choices for second digit} \times \text{Choices for third digit} = 2 \times 2 \times 1 = 4 \]

4. **Total number of even four-digit numbers**:
- Now, we add the results from both cases:
\[ \text{Total} = \text{Case 1} + \text{Case 2} = 6 + 4 = 10 \]

Thus, the total number of four-digit even numbers that can be formed using the digits 0, 1, 2, and 3 without repetition is **10**.