Let's consider the cases. case 1) When the numbers written on all three vertices are even: (2,2,2),(4,4,4),(2,2,4),(2,4,4) --->4x1=4. case 2) When two of the numbers written on the three vertices are even and the remaining one is odd: (2,2,1),(2,2,3),(2,2,5),(4,4,1),(4,4,3),(4,4,5),(2,4,1),(2,4,3),(2,4,5) The ordered pairs marked in red should consider circular permutations. --->6x1+3C1 x2!=6+6=12. case 3) When one of the numbers written on the three vertices is even and the other two are odd: A place where the product of two numbers is odd will occur. case 4) When the numbers written on all three vertices are odd: The product of two numbers becomes odd. Thus, the answer is 4+12=16.