How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?

To solve the problem of determining how many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 5, 6, and 7 with no repetition of digits, we need to consider the following conditions:

The number must be a 3-digit number.
The number must be even, meaning its last digit must be one of the even digits.
Digits cannot be repeated.
Step 1: Determine possible choices for the last digit
The last digit of the number must be even, so we can only use the digits 2, 4, or 6 from the given set {1, 2, 3, 4, 5, 6, 7}. Therefore, there are 3 choices for the last digit.

Step 2: Determine possible choices for the first digit
The first digit can be any of the remaining digits after selecting the last digit. Since the first digit must be non-zero, we can choose from the digits 1, 2, 3, 4, 5, 6, or 7, excluding the digit used for the last position. This leaves us with 6 available choices for the first digit.

Step 3: Determine possible choices for the second digit
The second digit can be any of the remaining digits after selecting the first and last digits. Since there are no restrictions on the second digit other than it must be different from the first and last digits, we have 5 choices left for the second digit.

Step 4: Calculate the total number of possible 3-digit even numbers
The total number of 3-digit even numbers can be calculated by multiplying the number of choices for each digit:

3 choices for the last digit (even digits: 2, 4, 6)
6 choices for the first digit
5 choices for the second digit
Therefore, the total number of 3-digit even numbers is: 3 * 6 * 5 = 90

Thus, the total number of 3-digit even numbers that can be formed is 90.