How many 4-digit even numbers can be formed using the digits 5, 6, 7 and 8 without repeating any digit?

Options

A. 12
B. 10
C. 11
D. 8

Explanation

To find how many 4-digit even numbers can be formed using the digits 5, 6, 7, and 8 without repeating any digit, we need to consider the properties of even numbers and the available digits.

### Step-by-Step Solution:

1. **Identify the Even Digits**:
The digits we have are 5, 6, 7, and 8. Among these, the even digits are 6 and 8. Therefore, the last digit of our 4-digit number must be either 6 or 8 to ensure the number is even.

2. **Case 1: Last Digit is 6**:
- If the last digit is 6, the remaining digits to choose from are 5, 7, and 8.
- We need to fill the first three positions with these three digits.
- The number of ways to arrange 3 digits (5, 7, and 8) in the first three positions is calculated by finding the factorial of the number of digits, which is \(3!\):
\[ 3! = 3 \times 2 \times 1 = 6 \]

3. **Case 2: Last Digit is 8**:
- If the last digit is 8, the remaining digits to choose from are 5, 6, and 7.
- Similarly, we need to fill the first three positions with these three digits.
- The number of ways to arrange 3 digits (5, 6, and 7) in the first three positions is also \(3!\):
\[ 3! = 3 \times 2 \times 1 = 6 \]

4. **Total Number of Even Numbers**:
- Now, we add the number of arrangements from both cases:
\[ \text{Total} = 6 \text{ (last digit 6)} + 6 \text{ (last digit 8)} = 12 \]

Thus, the total number of 4-digit even numbers that can be formed using the digits 5, 6, 7, and 8 without repeating any digit is **12**.