Step-by-step Solution:

The domain \( \{1, 2, 3, 4\} \) has 2 even numbers (2, 4) and 2 odd numbers (1, 3). The codomain \( \{1, 2, 3, 4, 5, 6, 7, 8\} \) has 4 even numbers and 4 odd numbers. The function \( f \) must be injective and send even numbers to even numbers. Step 1: Map the 2 even numbers from the domain to the 4 even numbers in the codomain. The number of ways is: \[ P(4, 2) = 4 \times 3 = 12 \] Step 2: Map the 2 odd numbers from the domain to any of the remaining 6 numbers in the codomain (since 2 numbers are already taken and the function is injective). The number of ways is: \[ P(6, 2) = 6 \times 5 = 30 \] Total number of injective functions \( n \) is: \[ n = 12 \times 30 = 360 \] Now, prime factorize 360: \[ 360 = 36 \times 10 = 2^3 \times 3^2 \times 5^1 \] So, \( a = 3, b = 2, c = 1 \). Therefore, \( a + b + c = 3 + 2 + 1 = 6 \).