counting probability with multiple cases

How many ways could you color the four different houses without any constraints?

Call this $N$.

Once you have that figured out, consider your one constraint re-stated as:

You are not allowed to use the same color for all four houses.

Note that there are only four cases in which your constraint is violated; i.e., all four houses color 1, all four houses color 2, all four houses color 3, and all four houses color 4.

So the total will be $N - 4$.

If you could paint any house with any color without restriction, there would be $4^4$ colors.

Now, note that you're looking at all colorings except for the four colorings which are all of one color. So you have $4^4 - 4$ ways to color the houses.

I think your way is ok. But easier would be to count all the ways to paint the $4$ houses (with no restrictions), and then subtract off the cases you don't want.