Question
How many different three-digit even integers can be made using the three digits 6,7 , and 8 ?
Ask by Barnett Bryant. in the United States
Mar 20,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
There are 4 different three-digit even integers that can be made using the digits 6, 7, and 8.
Solution
1. A three-digit even number must end with an even digit. Among the digits \(6\), \(7\), and \(8\), the even digits are \(6\) and \(8\).
2. We consider two cases based on the last digit:
- **Case 1: Last digit is \(6\)**
The remaining two positions (hundreds and tens) must be filled with the digits \(7\) and \(8\). There are \(2!\) ways to arrange these two digits:
\[
2! = 2
\]
- **Case 2: Last digit is \(8\)**
The remaining two positions (hundreds and tens) must be filled with the digits \(6\) and \(7\). There are also \(2!\) ways:
\[
2! = 2
\]
3. Adding the two cases:
\[
2 + 2 = 4
\]
Thus, there are \(\boxed{4}\) different three-digit even integers that can be made using the digits \(6\), \(7\), and \(8\).
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Simplify this solution The Deep Dive
To form a three-digit even integer using the digits 6, 7, and 8, the last digit must be an even number. Since 6 and 8 are the even digits available, we consider two cases based on the last digit being 6 or 8. 1. **Last digit is 6**: The first two digits can be either 7 or 8. The combinations are: - 786 - 876 So, there are 2 options for this case. 2. **Last digit is 8**: The first two digits can be either 6 or 7. The combinations are: - 678 - 768 Therefore, there are 2 options for this case as well. Adding both cases together, we find that there are a total of 2 + 2 = 4 different three-digit even integers that can be formed using 6, 7, and 8.
