To form a 3-digit number using the digits $0, 1, 2, 3, 4$ without repetition, we must consider the following:

Step 1: First Digit (Hundreds Place)

The first digit of a 3-digit number cannot be $0$, as that would make it a 2-digit number. So, the choices for the first digit are $1, 2, 3, 4$ (4 choices).

Step 2: Second Digit (Tens Place)

Once the first digit is chosen, we have 4 remaining digits available. Since repetition is not allowed, the number of choices for the second digit is now reduced to 4.

Step 3: Third Digit (Units Place)

After selecting the first two digits, we are left with 3 unused digits. Therefore, we have 3 choices for the third digit.

Step 4: Total Number of Combinations

The total number of 3-digit numbers is the product of the choices available for each place:

$\text{Total Ways} = 4 \times 4 \times 3 = 48$

Thus, the total number of 3-digit numbers that can be formed using the digits $0, 1, 2, 3, 4$ without repetition is 48.

Would you like more details on this solution, or do you have any further questions?

Here are 5 related questions to expand on this topic:

1. How many 4-digit numbers can be formed using the digits 0, 1, 2, 3, 4 without repetition?
2. What is the total number of ways to arrange all the digits $0, 1, 2, 3, 4$ into a 5-digit number?
3. How many 3-digit numbers can be formed using only even digits from $0, 1, 2, 3, 4$?
4. If repetition were allowed, how many 3-digit numbers could be formed from $0, 1, 2, 3, 4$?
5. What is the probability of forming a 3-digit number divisible by 5 using the digits $0, 1, 2, 3, 4$ without repetition?

Tip: When dealing with digit problems, always pay attention to whether repetition is allowed or not, as it greatly impacts the total number of possibilities.