To solve the problem of how many four-digit even integers can be formed using the digits 0, 1, 2, 3, 4, and 5, we will consider two cases: when digits can be repeated and when digits cannot be repeated.
### Case 1: Digits can be repeated
1. **Identify the last digit**: Since we need to form an even number, the last digit can only be 0, 2, or 4. This gives us **3 choices** for the last digit.
**Choices for the last digit**: 0, 2, 4 → **3 choices**
2. **Identify the first digit**: The first digit cannot be 0 (as it would not be a four-digit number). Therefore, if the last digit is 0, we can choose from 1, 2, 3, 4, and 5. If the last digit is 2 or 4, we can choose from 1, 2, 3, 4, and 5, excluding the last digit. This gives us **5 choices** for the first digit.
**Choices for the first digit**: 1, 2, 3, 4, 5 → **5 choices**
3. **Identify the second and third digits**: Both the second and third digits can be any of the 6 digits (0, 1, 2, 3, 4, 5) since repetition is allowed. Thus, we have **6 choices** for each of these digits.
**Choices for the second digit**: 0, 1, 2, 3, 4, 5 → **6 choices**
**Choices for the third digit**: 0, 1, 2, 3, 4, 5 → **6 choices**
4. **Calculate the total number of combinations**: The total number of four-digit even integers when digits can be repeated is given by multiplying the choices for each digit position.
\[
\text{Total} = (\text{Choices for first digit}) \times (\text{Choices for second digit}) \times (\text{Choices for third digit}) \times (\text{Choices for last digit})
\]
\[
\text{Total} = 5 \times 6 \times 6 \times 3 = 540
\]
### Case 2: Digits cannot be repeated
1. **Identify the last digit**: Again, the last digit can only be 0, 2, or 4. This gives us **3 choices** for the last digit.
2. **Identify the first digit**: The first digit cannot be 0. If the last digit is 0, the first digit can be any of 1, 2, 3, 4, or 5 (5 choices). If the last digit is 2 or 4, we can choose from 1, 3, 4, and 5 (4 choices).
- If last digit is 0: 5 choices
- If last digit is 2: 4 choices
- If last digit is 4: 4 choices
3. **Identify the second and third digits**: After choosing the last and first digits, we have to select the second and third digits from the remaining available digits.
- If the last digit is 0:
- First digit has 5 choices, leaving 4 digits for the second digit and 3 for the third digit.
- Total = \(5 \times 4 \times 3 = 60\)
- If the last digit is 2 or 4:
- First digit has 4 choices, leaving 4 digits for the second digit and 3 for the third digit.
- Total = \(4 \times 4 \times 3 = 48\) for each case (2 and 4).
4. **Calculate the total number of combinations**:
- Total for last digit 0: 60
- Total for last digit 2: 48
- Total for last digit 4: 48
\[
\text{Total} = 60 + 48 + 48 = 156
\]
### Final Answer
Now, we combine the results from both cases:
- Case 1 (with repetition): 540
- Case 2 (without repetition): 156
Thus, the total number of four-digit even integers that can be formed is:
\[
\text{Total} = 540 + 156 = 696
\]
