Interesting problem indeed: I am keen to try and see what is possible to fix down.
Premise
So we are considering the words, formed from the alphabet
$$
\alpha = \left\{ {1,2,\, \cdots ,\,n} \right\}
$$
with total number of repetitions of each character strictly correspondending to the vector
$$
{\bf r} = \left( {r_{\,1} ,\,r_{\,2} ,\, \cdots ,\,r_{\,n} } \right)
$$
and thus having length equal to
$$
R = \sum\limits_{1 \le \,k\, \le \,n} {r_{\,k} }
$$
With $r_k=0$ we will indicate that there is no occurence of the character $k$, which is a case
useful to consider in the following.
Of course, there is no loss of generality in permuting the vector $\bf r$, so that we may assume that its
components are arranged ,e.g., as non-increasing; then $a$ and $b$ shall also be permuted accordingly.
So
$$
{\bf r}_{\,{\bf n}} = \left( {r_{\,1} ,r_{\,2} ,\, \cdots ,\,r_{\,m} ,0,\, \cdots ,0} \right)\quad \Rightarrow \quad {\bf r}_{\,{\bf m}} = \left( {r_{\,1} ,r_{\,2} ,\, \cdots ,\,r_{\,m} } \right)
$$
The total number of words that can be formed from a given $\bf r$ (the alphabet is implicit in its dimension), by definition of Multinomial is:
$$
N_w ({\bf r}) = \left( \matrix{
R \cr
r_{\,1} ,r_{\,2} ,\, \cdots ,\,r_{\,n} \cr} \right) = {{R!} \over {r_{\,1} !r_{\,2} !\, \cdots \,r_{\,n} !}}
$$
Our focus is on the words that do not have equal contiguous characters.
The Multinomial approach
Consider the words in which one specific character (wlog, the first) does not appear contiguously,
i.e. in runs of not more than $1$ in length, while the others can be placed in whatever way.
Then, as explained in this post, their number will be:
$$ \bbox[lightyellow] {
\eqalign{
& N_{\,w1} (r_{\,1} ,R) = \left( \matrix{ R - r_{\,1} \cr
r_{\,2} ,\, \cdots ,\,r_{\,n} \cr} \right)N_b (r_{\,1} ,1,R - r_{\,1} + 1) = \left( \matrix{
R - r_{\,1} \cr
r_{\,2} ,\, \cdots ,\,r_{\,n} \cr} \right)\sum\limits_{\left( {0\, \le } \right)\,\,j\,\,\left( { \le \,r_{\,1} \,/2} \right)} {\left( { - 1} \right)^j \left( \matrix{
R - r_{\,1} + 1 \cr j \cr} \right)\left( \matrix{ R - 2j \cr r_{\,1} - 2\,j \cr} \right)} = \cr
& = \left( \matrix{ R - r_{\,1} \cr r_{\,2} ,\, \cdots ,\,r_{\,n} \cr} \right)\left( \matrix{ R - r_{\,1} + 1 \cr r_{\,1} \cr} \right)
= \left( \matrix{ R \cr r_{\,1} ,\,r_{\,2} ,\, \cdots ,\,r_{\,n} \cr} \right)
\left( \matrix{ R - r_{\,1} + 1 \cr r_{\,1} \cr} \right)\;\mathop /\limits_{} \;
\left( \matrix{ R \cr r_{\,1} \cr} \right) \cr}
\tag{1} }$$
If the other characters have repetition $1$, then clearly that is the number of words with no equal contiguous character.
In this case $R=n+r_{1}-1$
$$ \bbox[lightyellow] {
\eqalign{
& N_{\,w1} \left( {r_{\,1} ,\,1,\, \cdots ,1} \right) = \left( \matrix{
R \cr
r_{\,1} ,1,\, \cdots ,\,1 \cr} \right)\left( \matrix{
R - r_{\,1} + 1 \cr
r_{\,1} \cr} \right)\;\mathop /\limits_{} \;\left( \matrix{
R \cr
r_{\,1} \cr} \right) = \cr
& = \left( {R - r_{\,1} } \right)!\left( \matrix{
R - r_{\,1} + 1 \cr
r_{\,1} \cr} \right) = \left( {n - 1} \right)!\left( \matrix{
n \cr
r_{\,1} \cr} \right) \cr}
\tag{2} }$$
For the examples you give
$$
\eqalign{
& a,a,b\quad \Rightarrow \;\quad N_{\,w1} = 1!\left( \matrix{
2 \cr
2 \cr} \right) = 1 \cr
& a,a,b,c,d,e,f\quad \Rightarrow \;\quad N_{\,w1} = 5!\left( \matrix{
6 \cr
2 \cr} \right) = 1800 \cr}
$$
so, with your method of counting, you just have to multiply by $r_{1}!$.
For the case of a general $\bf {r}$, we could develop forward identity (1) and proceed by inclusion-exclusion, but looks cumbersome.
Let's try instead to find an adequate recurrence.
The recursive approach
Definitions
Let's call
$$
N_{\,0} ({\bf r},a,b)
$$
the N. of words with no equal contiguous character, with repetition vector $\bf r$ (the alphabet is implicit in it)
and with starting character $a$ and ending character $b$.
We consider empty, and thus their number null, the words for which:
- the starting and ending character do not pertain to the alphabet : $a \notin \alpha \,\; \vee \;b \notin \alpha $
- any component of the repetition vector is negative : $\exists k:r_{\,k} < 0$
- the start or the end character have null repetition : $r_{\,a} = 0\,\; \vee \;r_{\,b} = 0$
- the dimension of $\bf r$, i.e. of the alphabet, is null : $n<1$
In this way, it is clear that the words with a given starting and ending character constitute a partition
of the set of the considered words.
Let's then indicate
$$
N_{\,0} ({\bf r}) = \sum\limits_{1 \le \,k,\,j\, \le \,n} {N_{\,0} ({\bf r},k,j)}
$$
and put by definition that the number of empty words is 1
$$
N_{\,0} ({\bf 0}) = 1
$$
Also, we define the following vectors
$$
\eqalign{
& {\bf u}_{\,n} (a) = \left( {\delta _{\,k,\,a} \;\left| {\,1 \le k \le n} \right.} \right) = \left( {\left[ {a = k} \right]\;\;\left| {\,1 \le k \le n} \right.} \right) \cr
& {\bf u}_{\,n} (a,b) = \left( {\delta _{\,k,\,a} + \delta _{\,k,\,b} \;\left| {\,1 \le k \le n} \right.} \right) = \left( {\left[ {a = k} \right] + \left[ {b = k} \right]\;\;\left| {\,1 \le k \le n} \right.} \right) \cr
& {\bf u}_{\,n} = \left( {1\;\;\left| {\,1 \le k \le n} \right.} \right) \cr}
$$
where $\delta _{\,k,\,a} $ denotes the Kronecker delta and
$[P]$ denotes the Iverson bracket.
The starting cases
Then, considering the simplest case of $R=1$, we have
$$
N_{\,0} ({\bf u}_{\,n} (c),a,b) = \left[ {a = b = c} \right] = \left[ {a = c} \right]\left[ {b = c} \right]
$$
while for $R=2$
$$
N_{\,0} ({\bf u}_{\,n} (c,d),a,b) = \left[ {c \ne d} \right]\left( {\left[ {a = c} \right]\left[ {b = d} \right] + \left[ {a = d} \right]\left[ {b = c} \right]} \right)
$$
and for $R=3$
$$
N_{\,0} ({\bf u}_{\,n} (c,d,e),a,b) = \left[ {a \ne b} \right]\left[ {\left( {a,b} \right) \in 2{\rm - permutation}\,{\rm from}\,(c,d,e)} \right]
$$
and finally for ${\bf u}_{\,n} $ ($R=n$)
$$
N_{\,0} ({\bf u}_{\,n} ,a,b) = \left[ {1 = n} \right]\left[ {a = b} \right] + \left[ {2 \le n} \right]\left[ {a \ne b} \right]\left( {n - 2} \right)!
$$
The recursion
Take a word (with no equal contiguous character) of length $2 \le R$.
We can divide it into two subwords of fixed length $R-S$ and $1 \le S < R$.
The end character of the first shall be different from the starting character of the second.
Then we can part the repetition vector between them in all the ways, such that
the sum of the components of the first is $R-S$ and that of the second is $S$.
That means that $\bf{r}$ is reparted into $\bf{r-s}$ and $\bf{s}$.
Both vectors shall have at least one positive component. However, with the conditions
established above $N_{\,0} (\mathbf{0},a,b) = 0\quad \left| {\;1 \leqslant n} \right.$
We can therefore write $N_{\,0} (\mathbf{r},a,b)$ as the following sum in $j,k,\bf{s}$
$$ \bbox[lightyellow] {
\begin{gathered}
N_{\,0} (\mathbf{r},a,b) = \left[ {1 = R} \right]\left[ {a = b = k:r_{\,k} \ne 0} \right] + \hfill \\
+ \left[ {2 \leqslant R} \right]\sum\limits_{\left\{ \begin{subarray}{l}
1 \leqslant \,k \ne \,j\, \leqslant \,n \\
\mathbf{0}\, \leqslant \,\mathbf{s}\, \leqslant \,\mathbf{r}\, \\
0 < \,\sum\limits_l {s_{\,l} } = S\, < \,R
\end{subarray} \right.} {N_{\,0} (\mathbf{r} - \mathbf{s},a,k)\;N_{\,0} (\mathbf{s},j,b)} \hfill \\
\end{gathered}
\tag{3} }$$
Then, for $S=1$ , (and $2 \le R$), in particular, we have
$$ \bbox[lightyellow] {
\begin{gathered}
N_{\,0} (\mathbf{r},a,b) = \sum\limits_{\left\{ \begin{subarray}{l}
1 \leqslant \,k \ne \,j\, \leqslant \,n \\
\mathbf{0} \leqslant \,\mathbf{s}\, \leqslant \,\mathbf{r} \\
0 < \sum\limits_l {s_{\,l} } = 1\, < \,R
\end{subarray} \right.} {N_{\,0} (\mathbf{r} - \mathbf{s},a,k)N_{\,0} (\mathbf{s},j,b)} = \hfill \\
= \sum\limits_{\left\{ \begin{subarray}{l}
1 \leqslant \,k \ne \,j\, \leqslant \,n \\
1 \leqslant \,l\, \leqslant \,n \\
1 < \,R
\end{subarray} \right.} {N_{\,0} \left( {\left( {r_{\,1} ,\, \cdots ,r_{\,l} - 1,\, \cdots ,\,r_{\,n} } \right),a,k} \right)\;N_{\,0} \left( {\left( {0,\, \cdots ,1_{\,\left( l \right)} ,\, \cdots ,\,0} \right),j,b} \right)} = \hfill \\
= \sum\limits_{\left\{ \begin{subarray}{l}
1 \leqslant \,k \ne \,j\, \leqslant \,n \\
1 \leqslant \,l\, \leqslant \,n \\
1 < \,R
\end{subarray} \right.} {N_{\,0} (\left( {r_{\,1} ,\, \cdots ,r_{\,l} - 1,\, \cdots ,\,r_{\,n} } \right),a,k)\;\left[ {l = j = b} \right]} = \hfill \\
= \sum\limits_{\left\{ \begin{subarray}{l}
1 \leqslant \,k \ne \,b\, \leqslant \,n \\
1 < \,R
\end{subarray} \right.} {N_{\,0} (\left( {r_{\,1} ,\, \cdots ,r_{\,b} - 1,\, \cdots ,\,r_{\,n} } \right),a,k)} = \hfill \\
= \sum\limits_{1 \leqslant \,k \ne \,b\, \leqslant \,n} {N_{\,0} (\mathbf{r} - \mathbf{u}(b),a,k)} = \hfill \\
= \sum\limits_{1 \leqslant \,k\, \leqslant \,n} {N_{\,0} (\mathbf{r} - \mathbf{u}(b),a,k)} - N_{\,0} (\mathbf{r} - \mathbf{u}(b),a,b) \hfill \\
\end{gathered}
\tag{4} }$$
which I checked to be correct against a dozen of cases.
