How many ways to read the palindromic word ROTATOR in this design

algebra-precalculuscombinatoricsrecreational-mathematics

The problem is as follows:

The figure from below shows the word $\textrm{ROTATOR}$ arranged in a
peculiar way. How many ways can this word be read assuming the equal
least distance from one letter to another?.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{490 ways}\\
2.&\textrm{480 ways}\\
3.&\textrm{245 ways}\\
4.&\textrm{400 ways}\\
\end{array}$

I noticed that the word is palindromic, hence it can be read back and forth, thus this means that I should account for these possibilites.

In order to keep the right track for this purpose I used an auxiliary numbers atop the letters to account for these as shown in the diagram from below.

Sketch of the solution

After doing all of that I reached the conclusion that:

$\textrm{ways}=(74+74+96)\times 2 =488$

But this doesn't appear in any of the alternatives. Did I mess up something or what? Can someone help me here? Please I require a step by step explanation as I feel lost if my method did worked out properly?

Best Answer

Here is a slightly easier argument.
Let's just count how many ways there are to go from an R to an A. The R's on one side give this:

$$R^1 \quad R^1 \quad R^1\\ O^1 \quad O^2 \quad O^2 \quad O^1 \\ T^3 \quad T^4 \quad T^3 \\ A^3 \quad A^7 \quad A^7 \quad A^3$$

Starting at the R's on the other end gives this:

$$A^4 \quad A^7 \quad A^7 \quad A^4 \\ T^1 \quad T^3 \quad T^4 \quad T^3 \quad T^1 \\ O^1 \quad O^2 \quad O^2 \quad O^1 \\ R^1 \quad R^1 \quad R^1$$

Adding those together, there are $7$, $14$, $14$, and $7$ ways to go from any R to each of the central A's. Conversely there are the same number of ways to go from each of those A's back to any R. Combining any R-to-A path with any A-to-R path from the same A we get $7\cdot7+14\cdot14+14\cdot14+7\cdot7 = 490$.

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