We are tasked with finding how many 3-digit numbers can be constructed from the digits 0-9, using each digit at most once, with the condition that the number ends in 0.

Steps:

1. Fix the last digit: Since the number must end in 0, the last digit is fixed as 0.

2. Choose the first digit: The first digit must be a non-zero digit (since it’s a 3-digit number), and it must be chosen from the digits 1-9. So, there are 9 possible choices for the first digit.

3. Choose the second digit: After the first digit is chosen, one digit has been used up. Therefore, there are 8 remaining digits to choose from for the second digit.

Total number of possibilities:

• First digit: 9 choices (1-9),
• Second digit: 8 choices (remaining digits after choosing the first),
• Last digit: fixed as 0.

Thus, the total number of 3-digit numbers that end in 0 is:

$9 \times 8 = 72$

Therefore, 72 numbers can be formed where the 3-digit number ends in 0.

Would you like to see more details or ask any further questions?

Related questions:

1. How many 3-digit numbers can be formed without any restrictions?
2. How many 3-digit numbers can be formed that start with 0?
3. How many 3-digit numbers can be formed that do not contain the digit 0?
4. How many 4-digit numbers can be formed with distinct digits?
5. How many 3-digit numbers can be formed that end in a specific digit, say 5?

Tip:

When solving problems involving digits, always check if the placement of 0 is restricted since it cannot be the first digit of a multi-digit number.