<s>[INST] Number of $6$-digit numbers made up of the digits $1$, $2$, and $3$ with no digit occurring $3$ or more times consecutively? Find the number of 6-digit numbers made up of the digits $1$, $2$, and $3$ that have no digit occur three or more times consecutively. (For example, $123123$ would count, but $123111$ would not.)
We know the total number of possibilities without the occurrence restriction is $3^6 = 729$. 
In order to eliminate the possibilities that can't be used, I set a block of 3 numbers as my 3 consecutive digits block. This block would have 3 different combinations. I then split the number of eliminations into 2 cases: one where the block of 3 is in the front, and one where the block of 3 is in the middle. 
For case 1: I computed $3\cdot 24$, with $3$ being the ways to change the block of 3, and $24$ being the number of ways to change the other 3 numbers while not having another block of 3 consecutive. 
For case 2: I computed $3^4$, ignoring the restriction of case 1. 
Summing the 2 cases, I got $72+ 3\cdot 81 = 315$, I then subtracted from $729$ and got $414$, which turns out to be incorrect. What am I doing wrong?
 [/INST] Discussion of what's missing in your answer
There are four places a block of three can start in the number.
$$\begin{matrix}
(1): && d & d & d & \cdot & \cdot & \cdot  \\
(2): && \cdot & d & d & d & \cdot & \cdot  \\
(3): && \cdot & \cdot & d & d & d & \cdot  \\
(4): && \cdot & \cdot & \cdot & d & d & d  \\
\end{matrix}$$
You have only discussed the first two.  Additionally, if we just count these four configurations, we overcount configurations where there are four, five, or six consecutive digits.  For instance, $111111$ can only be generated one way, but it has been included in the counts for configurations (1), (2), (3), and (4), so has been quadruple-counted.
Corrected version of your answer
There are ten ways to have three (or more) consecutive digits: 
$$\begin{matrix}
(1): && d & d & d & \cdot & \cdot & \cdot  \\
(2): && \cdot & d & d & d & \cdot & \cdot  \\
(3): && \cdot & \cdot & d & d & d & \cdot  \\
(4): && \cdot & \cdot & \cdot & d & d & d  \\
(5): && d & d & d & d & \cdot & \cdot  \\
(6): && \cdot & d & d & d & d & \cdot  \\
(7): && \cdot & \cdot & d & d & d & d  \\
(8): && d & d & d & d & d & \cdot  \\
(9): && \cdot & d & d & d & d & d  \\
(10): && d & d & d & d & d & d
\end{matrix} $$
We will proceed by inclusion-exclusion.
And then proceed as shown by N.F.Taussig's answer.  (I was about half way through when he saved me having to finish it.)
A solution that avoids explicit inclusion-exclusion
From each string of six digits, $d_1 d_2 d_3 d_4 d_5 d_6$, let us construct a new string of length $5$ having an "s" in position $i$ if $d_i = d_{i+1}$ and an "n" if $d_i \neq d_{i+1}$, as $i$ ranges from $1$ to $5$.  A digit occurring three or more times consecutively corresponds to two or more consecutive "s"s.  There are only $3$ choices of first pairs of digits which produce an "s" in the first position and $6$ choices of pairs of digits which produce an "n" in the first position.  Following an "s" only one choice of digit can produce an "s" (which corresponds to disallowed three consecutive digits) and two choices of digit produce an "n".  Following an "n" one choice of digit produces an "s" and two choices of digit produce an "n".
The acceptable 5 letter strings, with their counts and a running sum, are \begin{align*}
snsns  &&  3 \cdot 2 \cdot 1 \cdot 2 \cdot 1 = 12  && 12  \\
snsnn  &&  3 \cdot 2 \cdot 1 \cdot 2 \cdot 2 = 24  && 36  \\
snnsn  &&  3 \cdot 2 \cdot 2 \cdot 1 \cdot 2 = 24  && 60  \\
snnns  &&  3 \cdot 2 \cdot 2 \cdot 2 \cdot 1 = 24  && 84  \\
snnnn  &&  3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 48  && 132  \\
nsnsn  &&  6 \cdot 1 \cdot 2 \cdot 1 \cdot 2 = 24  && 156  \\
nsnns  &&  6 \cdot 1 \cdot 2 \cdot 2 \cdot 1 = 24  && 180  \\
nsnnn  &&  6 \cdot 1 \cdot 2 \cdot 2 \cdot 2 = 48  && 228  \\
nnsns  &&  6 \cdot 2 \cdot 1 \cdot 2 \cdot 1 = 24  && 252  \\
nnsnn  &&  6 \cdot 2 \cdot 1 \cdot 2 \cdot 2 = 48  && 300  \\
nnnsn  &&  6 \cdot 2 \cdot 2 \cdot 1 \cdot 2 = 48  && 348  \\
nnnns  &&  6 \cdot 2 \cdot 2 \cdot 2 \cdot 1 = 48  && 396  \\
nnnnn  &&  6 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 96  && 492  \\
\end{align*}
So there are $492$ allowed configurations.
I generally prefer this method because it more directly corresponds to the tree of consecutive digit choices.
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