How many 3-digit even numbers can be formed with no digit repeated by using the digits 0, 1, 2, 3, 4 and 5?

Given:

We are asked to find how many 3-digit even numbers can be formed with no digit repeated using the digits 0, 1, 2, 3, 4, and 5.

Solution:

For a 3-digit number to be even, its unit (last) digit must be one of the even digits. In this case, the even digits available are 0, 2, and 4.

Case 1: The unit digit is 0

If the unit digit is 0, then the first digit (hundreds place) can be chosen from the remaining digits 1, 2, 3, 4, and 5. There are 5 choices for the first digit.

The second digit (tens place) can be chosen from the remaining 4 digits, giving us 4 choices for the second digit.

Thus, the total number of even 3-digit numbers when the unit digit is 0 is:

5 × 4 = 20

Case 2: The unit digit is 2

If the unit digit is 2, then the first digit (hundreds place) can be chosen from the remaining digits 1, 3, 4, and 5. There are 4 choices for the first digit.

The second digit (tens place) can be chosen from the remaining 4 digits (0, 1, 3, 4, 5), giving us 4 choices for the second digit.

Thus, the total number of even 3-digit numbers when the unit digit is 2 is:

4 × 4 = 16

Case 3: The unit digit is 4

If the unit digit is 4, then the first digit (hundreds place) can be chosen from the remaining digits 1, 2, 3, and 5. There are 4 choices for the first digit.

The second digit (tens place) can be chosen from the remaining 4 digits (0, 1, 2, 3, 5), giving us 4 choices for the second digit.

Thus, the total number of even 3-digit numbers when the unit digit is 4 is:

4 × 4 = 16

Total number of 3-digit even numbers:

20 (unit digit 0) + 16 (unit digit 2) + 16 (unit digit 4) = 52

Answer: 52