How many different three-digit even integers can be made using the three digits 6,7 , and 8 ?

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 \(8\).