Let $W(c,n)$ denote the number of words of length $c$ from an alphabet of $n$ letters. Then $W(c,n)=n^c$.
Out of these, the number of words of the same size that do not contain one of the letters is $W(c,n-1)=(n-1)^c$. The number of ways of choosing which letter is missing is $\binom{n}{1}$.
The number of words of the same size that do not contain two letters is $W(c,n-2)=(n-2)^c$. The number of ways of choosing which two letters are missing is $\binom{n}{2}$... and so on ...
Now we use inclusion-exclusion principle: (subtract the number of words missing one of the letters, then add the number missing two of the letters, subtract the number missing three of the letters,...)
We get:
$$W(c,n)-\binom{n}{1}W(c,n-1)+\binom{n}{2}W(c,n-2)-\binom{n}{3}W(c,n-3)+\cdots+(-1)^{n-1}\binom{n}{n-1}W(c,n-(n-1)).$$
This is
$$n^c-\binom{n}{1}(n-1)^c+\binom{n}{2}(n-2)^c-\binom{n}{3}(n-3)^c+\cdots+(-1)^{n-1}\binom{n}{n-1}1^c.$$
or
$$\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}(n-k)^c.$$
Another way could be: Denote $S_c^n$ the number of ways to partition the word of length $c$ into $n$ pieces. Then we just need to choose which letter goes to each of the $n$ pieces. This number is $n!$. So the number of words we are looking for is
$$n!S_c^n.$$
The numbers $S_c^n$ are called Stirling's numbers of the second kind.
Let $s_n$ count the admissible strings of lenth $n\geq0$, and $c_n$ the admissible strings of length $n$ not ending in $a$ or $b$. Then
$$s_0=c_0=1,\quad s_1=3\ .\tag{1}$$
If an admissible word ends in $a$ or $b$ there are two options for the next letter, otherwise three. Furthermore we can always append a $c$. It follows that we have
$$s_{n+1}=2s_n+c_n,\qquad c_{n+1}=s_n\ .$$
Eliminating $c_n$ we obtain the recursion
$$s_{n+1}=2s_n+s_{n-1}\qquad(n\geq1)\tag{2}$$
with characteristic polynomial $\lambda^2-2\lambda-1$. The roots are $1\pm\sqrt{2}$, so that the general solution of $(2)$ is
$$x_n=A\bigl(1+\sqrt{2}\bigr)^n + B\bigl(1-\sqrt{2}\bigr)^n\qquad(n\geq0)\ .$$
Using $(1)$ we then obtain
$$s_n={\bigl(1+\sqrt{2}\bigr)^{n+1} +\bigl(1-\sqrt{2}\bigr)^{n+1}\over2}\qquad(n\geq0)\ .\tag{3}$$
As the second term in $(3)$ is rapidly decreasing we may write
$$s_n={\rm round}\left({1\over2}\bigl(1+\sqrt{2}\bigr)^{n+1} \right)\qquad(n\geq0)\ .$$
Best Answer
Outline: Use Inclusion/Exclusion. Count first the number of words with a double A. It is perhaps easy to see that there are $\frac{7!}{2!2!2!}$ such words. Similarly, there are $\frac{7!}{2!2!2!}$ with a double B, and so on. So our first estimate is that there are $4\cdot \frac{7!}{2!2!2!}$ words of the desired kind.
However, this double-counts the number of words with, for example, a double A and a double B. There are $\frac{6!}{2!2!}$ such words. There are $\binom{4}{2}$ ways to choose the pair that we will have two doubles of. So our next estimate of the required number of words is $4\cdot \frac{7!}{2!2!2!}-\binom{4}{2}\cdot \frac{6!}{2!2!}$.
However, we have subtracted too much, for we have subtracted one too many times the $\binom{4}{3}\cdot \frac{5!}{2!}$ words of that have three doubles.
After we have added back $\binom{4}{3}\cdot \frac{5!}{2!}$, there is only a little adjustment left to make.