I believe I'm overthinking this or otherwise confused but I believe that the method to solve this would be $2^n$ where n is the length of the bytes? So in this particular case it would be $2^8$ equal 256 possible.
But then I feel like that isn't right and I'm mixed up. What I thought about is that there is 4 possible ways to have an even number f zeros (i.e. 2 zeros, 4 zeros, 6 zeros, or 8 zeros).
Any insight would be awesome as I'm confused...