# Number of Ternary Strings without a 2 followed by a 0

This is a problem from my textbook:

Call a ternary string good if it never contains a $$2$$ followed immediately by a $$0$$; otherwise, call it bad. Let $$g(n)$$ be the number of good strings of length $$n$$. Obviously $$g(1)=3$$, since all strings of length $$1$$ are good. Also, $$g(2)=8$$ since the only bad string of length $$2$$ is $$(2,0)$$. Now consider a value of $$n$$ larger than $$2$$. Write a recursive formula for $$g(n)$$.

The textbook provides the solution but I am confused about it. It says that if the last digit is a $$0$$, then there are $$g(n-1)$$ valid strings that can precede it, and of these $$g(n-1)$$, $$g(n-2)$$ end in a $$2$$. I'm struggling with the intuition of why there are $$g(n-2)$$ strings ending in a $$2$$. Sorry if this is an obvious question, I'm just really confused.

You can think of "ending at $$2$$" as a process of adding $$2$$ to the last of the original string whose length is $$n-2$$. Apparently during this process, no extra bad strings are gonna pop out.
Therefore, the number of good strings (with length $$n-1$$) ending in a $$2$$ should equal $$g(n-2)$$.