There is a fairly systematic way to work this type of question. You were specifically wondering about:
How many 4-letter words can be formed from the letters: ABBCCC?
First, you write the 'source partition' for your word:
[C,B,A]
[3,2,1] <-- this is the `source partition`
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.
[3,1] requests one triple and 1 single
[2,2] requests 2 doubles
[2,1,1] requests one double and 2 singles
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:
CCBC # type [3,1]
The expression for that type is:
nCr(1,1)*nCr(2,1) * fac(4)/fac(3)
'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. Hence nCr(2,1) The call to fac(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:
# The `source partition` for the word 'ABBCCC' is:
# [3,2,1]
# # target
w = 0 # ------
w += nCr(1,1)*nCr(2,1) * fac(4)/fac(3) # [3,1]
w += nCr(2,2) * fac(4)/(fac(2)*fac(2)) # [2,2]
w += nCr(2,1)*nCr(2,2) * fac(4)/fac(2) # [2,1,1]
#
# The answer is 38
Your answer is off by a factor of $2$, seems that you forgot the double letter L.
The word GOOGOLPLEX consists of the letters O (3x), G (2x), L (2x), P (1x), E (1x) and X (1x).
So the number of possible words consisting of these letters is the multinomial coefficient
$$
\binom{10}{3,2,2,1,1,1} = \frac{10!}{3!\cdot 2!\cdot 2!\cdot 1!\cdot 1!\cdot 1!} = 151200.
$$
Best Answer
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.