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?

There are $3^6 = 729$ possible sequences.  From these, we must subtract those in which three consecutive digits are the same.
Observe that a prohibited sequence must begin in one of the first four positions.  Let $A_1, A_2, A_3, A_4$ be the set of outcomes in which three consecutive digits beginning in the first, second, third, and fourth positions, respectively, are the same.
The set of outcomes that violate the restriction is $A_1 \cup A_2 \cup A_3 \cup A_4$.  By the Inclusion-Exclusion Principle, 
\begin{align*}
|A_1 \cup A_2 \cup A_3 \cup A_4| & = |A_1| + |A_2| + |A_3| + |A_4|\\
                                 & \quad - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_1 \cap A_4| - |A_2 \cap A_4| - |A_3 \cap A_4|\\ 
                                 & \quad + |A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_4|\\
                                 & \quad - |A_1 \cap A_2 \cap A_3 \cap A_4|
\end{align*}
$|A_1|$:  There are three ways of selecting the number that occupies the first three positions and three ways of selecting the number that occupy each of the last three positions.  Hence, $|A_1| = 3^4$.  By symmetry,
$$|A_1| = |A_2| = |A_3| = |A_4| = 3^4$$
$|A_1 \cap A_2|$:  Since the first three and the second three positions overlap, the first four positions must be filled with the same number.  There are three ways to select this number and three ways to fill each of the remaining two positions.  Hence, $|A_1 \cap A_2| = 3^3$.  By symmetry,
$$|A_1 \cap A_2| = |A_2 \cap A_3| = |A_3 \cap A_4| = 3^3$$
$|A_1 \cap A_3|$:  Since the first three and the third three positions overlap, the first five positions must be filled with the same number.  There are three ways to select this number and three ways to fill the remaining position.  Hence, $|A_1 \cap A_3| = 3^2$.  By symmetry,
$$|A_1 \cap A_3| = |A_2 \cap A_4| = 3^2$$
$|A_1 \cap A_4|$:  Since the first three and last three positions do not overlap, the first three positions can be filled in three ways, as can the last three.  Hence, 
$$|A_1 \cap A_4| = 3^2$$
$|A_1 \cap A_2 \cap A_3|$:  Since these sets overlap, the first five positions must be filled with the same number.  There are three ways to choose this number and three ways to fill the remaining position.  Thus, $|A_1 \cap A_3 \cap A_3| = 3^2$.  By symmetry,
$$|A_1 \cap A_2 \cap A_3| = |A_2 \cap A_3 \cap A_4| = 3^2$$
$|A_1 \cap A_2 \cap A_4|$:  Since the first three positions overlap with the second three and the second three positions overlap with the last three positions, all six positions must be the same.  There are three ways to select the number that fills all six positions.  Hence, $|A_1 \cap A_2 \cap A_4| = 3$. By symmetry,
$$|A_1 \cap A_2 \cap A_4| = |A_1 \cap A_3 \cap A_4| = 3$$
$|A_1 \cap A_2 \cap A_3 \cap A_4|$:  All six positions must be filled with the same number.  Since there are three ways to select the number,
$$|A_1 \cap A_2 \cap A_3 \cap A_4| = 3$$
By the Inclusion-Exclusion Principle,
\begin{align*}
|A_1 \cup A_2 \cup A_3 \cup A_4| & = 3^4 + 3^4 + 3^4 + 3^4\\
                                 & \quad - 3^3 - 3^2 - 3^2 - 3^3 - 3^2 - 3^3\\
                                 & \quad + 3^2 + 3 + 3 + 3^2\\
                                 & \quad - 3\\ 
                                 & = 237
\end{align*}
Hence, there are $729 - 237 = 492$ admissible sequences.
