Let $o(n)$ be the number of such words of length $n$ with an odd number of $A$’s. We can build these words by choosing an odd $k\in\{0,1,\ldots,n\}$, picking $k$ positions in the word for $A$’s, and filling the other $n-k$ positions with letters from the set $\{B,C,D,E,F\}$; given $k$, this can be done in $\binom{n}k5^{n-k}$ ways, so
$$o(n)=\sum_{{k=0}\atop{k\text{ odd}}}^n\binom{n}k5^{n-k}\;.$$
Let $e(n)$ be the number of words of length $n$ with an even number of $A$’s; clearly
$$e(n)=\sum_{{k=0}\atop{k\text{ even}}}^n\binom{n}k5^{n-k}\;.$$
There are $6^n$ words of length $n$, so $$e(n)+o(n)=6^n\;.\tag{1}$$ On the other hand,
$$\begin{align*}
e(n)-o(n)&=\sum_{{k=0}\atop{k\text{ even}}}^n\binom{n}k5^{n-k}-\sum_{{k=0}\atop{k\text{ odd}}}^n\binom{n}k5^{n-k}\\\\
&=\sum_{k=0}^n\binom{n}k(-1)^k5^{n-k}\\\\
&=(-1+5)^n\\\\
&=4^n\;.
\end{align*}\tag{2}$$
Combine $(1)$ and $(2)$ to solve for $o(n)$ and $e(n)$.
Added: You can also approach it by finding a recurrence and solving it for a closed form. Let $O_n$ be the set of words of length $n$ with an odd number of $A$’s, and let $E_n$ be the set of words of length $n$ with an even number of $A$’s. Let $w\in O_n$, let $w_f$ and $w_\ell$ be respectively the first and last letters of $w$, and let $w'$ be the word of length $n-2$ obtained by deleting $w_f$ and $w_\ell$ from $w$. If exactly one of $w_f$ and $w_\ell$ is an $A$, then $w'\in E_{n-2}$; otherwise, $w'\in O_{n-2}$. Conversely, any $w'\in E_{n-2}$ can be extended to a word in $O_n$ in $10$ ways by adding an $A$ at one end and a non-$A$ at the other, and any $w'\in O_{n-2}$ can be extended to a word in $O_n$ in $26$ ways by appending an $A$ at each end or a non-$A$ at each end. It follows that
$$\begin{align*}
o(n)&=10e(n-2)+26o(n-2)\\
&=10\big(o(n-2)+e(n-2)\big)+16o(n-2)\\
&=10\cdot6^{n-2}+16o(n-2)\;.
\end{align*}$$
This recurrence can be solved by a variety of techniques.
Let us look for example at $4$-letter words. They are of two types: (i) words with at most one $a$ and (ii) words with two $a$'s.
For Type (i), there are $4!$. For Type (ii), we choose the location of the $a$'s in $\binom{4}{2}$ ways. For each such way, the remaining two places can be filled in $(3)(2)$ ways.
The same strategy works for $3$-letter words. There are $(4)(3)(2)$ ways to have at most one $a$. For the words with two $a$'s, the locations can be picked in $\binom{3}{2}$ ways, and for each such way the remaining slot can be filled in $3$ ways.
Note that if we had more non-duplicated letters, the analysis would be much the same.
Best Answer
There is a fairly systematic way to work this type of question. You were specifically wondering about:
First, you write the 'source partition' for your word:
Note that the source partition provides for 1 triple, 1 double, and 1 single. Corresponding to that, you write every possible 'target partition' of 4-letters.
For each 'target partition', you write an expression that gives the number of ways that the given target type can occur. An example of the target type [3,1] is:
The expression for that type is:
'nCr( n, r)' is the function for 'combinations', which gives the number of ways you can make a unique selection from n distinct things, taking r at a time. 'fac( n)' is the function for the factorial of n.
Note that source [3,2,1] provides 1 triple, and target [3,1] requests 1 triple. Hence
nCr(1,1). After the triple is used up, the source [3,2,1] can provide 2 singles, and the target [3,1] requests 1 single. HencenCr(2,1)The call tofac(4), in each expression, always corresponds to the 4-letter word. Any division, if it occurs in an expression, corresponds to each multiple request of the corresponding target partition. That's all there is to the method, but it isn't always easy to get every last detail correct. The entire computation, as I did it in the programming language Python, follows: