To solve the problem of how many 5-digit and 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 without repetition, we will break it down into two parts.
### Part (i): Finding the number of 5-digit numbers
1. **Identify the digits available**: We have the digits 1, 2, 3, 4, and 5.
2. **Determine the number of choices for each digit position**:
- For the **1st position**, we can choose any of the 5 digits (1, 2, 3, 4, 5). So, we have **5 choices**.
- For the **2nd position**, we can choose from the remaining 4 digits (since we cannot repeat the digit used in the 1st position). So, we have **4 choices**.
- For the **3rd position**, we can choose from the remaining 3 digits. So, we have **3 choices**.
- For the **4th position**, we can choose from the remaining 2 digits. So, we have **2 choices**.
- For the **5th position**, we have only 1 digit left. So, we have **1 choice**.
3. **Calculate the total number of 5-digit combinations**:
- Total combinations = 5 × 4 × 3 × 2 × 1 = 120.
### Part (ii): Finding the number of 3-digit numbers
1. **Identify the digits available**: We still have the digits 1, 2, 3, 4, and 5.
2. **Determine the number of choices for each digit position**:
- For the **1st position**, we can choose any of the 5 digits. So, we have **5 choices**.
- For the **2nd position**, we can choose from the remaining 4 digits. So, we have **4 choices**.
- For the **3rd position**, we can choose from the remaining 3 digits. So, we have **3 choices**.
3. **Calculate the total number of 3-digit combinations**:
- Total combinations = 5 × 4 × 3 = 60.
### Final Answers:
- The number of 5-digit numbers that can be formed is **120**.
- The number of 3-digit numbers that can be formed is **60**.
