My answer is 237 and my approach is not vary smart. would like to know if they are more wise method for it. Here is my thinking: $3^5= 243. $ Then minus 15, since
(1) 0 0 0 0 or 0 0 0 0 (1),
(2) 0 0 0 0 or 0 0 0 0 (2), 0 0 0 0 0,
(0) 1 1 1 1 or 1 1 1 1 (0),
(2) 1 1 1 1 or 1 1 1 1 (2), 1 1 1 1 1,
(0) 2 2 2 2 or 2 2 2 2 (0),
(1) 2 2 2 2 or 2 2 2 2 (1),
(0) 2 2 2 2 or 2 2 2 2 (0), 2 2 2 2 2,
Wondering if there is any formula to calculate more systematically?
On
Let us count the number of strings where one symbol, call it $a$ appears as $aaaa$ in the string. You have $5$ places to fill to make a complete string, out of which $4$ (consecutive positions) will be occupied by the repeated symbol ($a$). So your string will be of the form $Xaaaa$ or $aaaaX$. In each scenario, the blank space can be filled in three ways by either $a$ or $b$ or $c$. So in all $6$ possibilities. However, when $a$ is filled in the blank space, then we get the same string $\color{red}{a}aaaa=aaaa\color{red}{a}$ and we have counted it twice. So in all $5$ strings, will be there where a given symbol has appeared in $4$ consecutive positions. Now for each symbol you can do this, so in all $15$ strings with this property.
General problem: If you are given $n$ symbols to make strings of length $L$ such that one symbol appears in $L-1$ consecutive positions. Then the number of such strings are given by $n(2n-1)$.
I don't think you will get more systematic, except that you might do that counting in fewer steps by saying in pseudocode logic if $(x \neq A)$ $\{xAAAA~\text{or}~AAAAx\}~\text{else}~\{xAAAA\}$ where $x,A \in \{0,1,2\}$ implies a count of $2 \cdot 3 \cdot 2 + 1 \cdot 3 \cdot 1 = 15$.
The advantage here is we can compute the number if our digits are a higher base. As is, it isn't definitively simpler.