Counting words with parity restrictions on the letters Let $a_n$ be the number of words of length $n$ from the alphabet $\{A,B,C,D,E,F\}$ in which $A$ appears an even number of times and $B$ appears an odd number of times.
Using generating functions I was able to prove that $$a_n=\frac{6^n-2^n}{4}\;.$$
I was wondering if the above answer is correct and in that case what could be a combinatorial proof of that formula?

I don’t know whether you’d call them combinatorial, but here are two completely elementary arguments of a kind that I’ve presented in a sophomore-level discrete math course. Added: Neither, of course, is as nice as Didier’s, which I’d not seen when I posted this.
Let $b_n$ be the number of words of length $n$ with an odd number of $A$’s and an odd number of $B$’s, $c_n$ the number of words of length $n$ with an even number of $A$’s and an even number of $B$’s, and $d_n$ the number of words of length $n$ with an odd number of $A$’s and an even number of $B$’s. Then
$$a_{n+1}=4a_n+b_n+c_n\;.\tag{1}$$
Clearly $a_n=d_n$: interchanging $A$’s and $B$’s gives a bijection between the two types of word. $(1)$ therefore reduces to
$$a_{n+1}=4a_n+b_n+c_n\;.\tag{2}$$
But $6^n=a_n+b_n+c_n+d_n=b_n+c_n+2a_n$, so $b_n+c_n=6^n-2a_n$, and $(2)$ becomes $$a_{n+1}=2a_n+6^n,a_0=0\;.\tag{3}$$ $(3)$ can be solved by brute force:
$$\begin{align*}
a_n&=2a_{n-1}+6^{n-1}\\
&=2(2a_{n-2}+6^{n-2})+6^{n-1}\\
&=2^2a_{n-2}+2\cdot 6^{n-2}+6^{n-1}\\
&=2^2(2a_{n-3}+6^{n-3})+2\cdot 6^{n-2}+6^{n-1}\\
&=2^3a_{n-3}+2^2\cdot 6^{n-3}+2\cdot 6^{n-2}+6^{n-1}\\
&\;\vdots\\
&=2^na_0+\sum_{k=0}^{n-1}2^k6^{n-1-k}\\
&=6^{n-1}\sum_{k=0}^{n-1}\left(\frac13\right)^k\\
&=6^{n-1}\frac{1-(1/3)^n}{2/3}\\
&=\frac{2^{n-1}3^n-2^{n-1}}{2}\\
&=\frac{6^n-2^n}4\;.
\end{align*}$$
Alternatively, with perhaps just a little more cleverness $(1)$ can be expanded to
$$\begin{align*}
a_{n+1}&=4a_n+b_n+c_n\\
b_{n+1}&=4b_n+2a_n\\
c_{n+1}&=4c_n+2a_n\;,
\end{align*}\tag{4}$$
whence 
$$\begin{align*}a_{n+1}&=4a_n+4b_{n-1}+2a_{n-1}+4c_{n-1}+2a_{n-1}\\
&=4a_n+4a_{n-1}+4(b_{n-1}+c_{n-1})\\
&=4a_n+4a_{n-1}+4(a_n-4a_{n-1})\\
&=8a_n-12a_{n-2}\;.
\end{align*}$$
This straightforward homogeneous linear recurrence (with initial conditions $a_0=0,a_1=1$) immediately yields the desired result.
