Oozy rat in a sanitary zoo

We will first count all $\text{SANITARYZOO}$ we can make. If we focus on each quarter of the diamond grid, we find that there is 1 way to choose the $\text{S}$ , after which there are two choices for each letter i.e. there are 2 ways to choose an $\text{A}$ , after which there are two ways of choosing an $\text{N}$ and so forth. Since there are 11 letters in $\text{SANITARYZOO}$ , there are $2^{10}$ of these in each quarter. Thus, there are $4 \times 2^{10}$ altogether. However, we have over-counted when the words lie along the vertical or horizontal diagonals. Thus, there are actually $4\times 2^{10} - 4$ words. Since each of these words can be paired with any other (including itself) to create $\text{OOZYRATINASANITARYZOO}$ , there are $(4\times 2^{10} - 4)^2=1020^2=16744474$ total words.

Therefore, there are $16744464$ total ways of reading $\text{OOZYRATINASANITARYZOO}$