(Exponential) generating functions may well be the best approach. Unfortunately, I don't know anything about generating functions. So, I can only offer the following alternative, extremely crude approach.
Unfortunately, the approach taken is so convoluted, that for someone relatively new to the topic, giving a helpful hint would be very problematic. So, I have given a full solution.
The first thing to do is to identify how many distinct satisfying combinations of letters that there are. Then, for each such combination, you would need to apply a scaling factor.
For example, one combination would be a $5$ letter word, consisting of $2$ a's and $3$ b's. This specific combination can be used to create $\binom{5}{2}$ words, since there are $\binom{5}{2}$ ways that the $2$ a's can be positioned.
In general, if you have an $n$ letters consisting of $r$ a's, $s$ b's and $t$ c's (i.e. where $r + s + t = n$), then the number of words that can be formed from this particular combination (i.e. the scaling factor) is
$$\frac{n!}{r! \times s! \times t!}.\tag1$$
To start, I suggest that you read the answer to this question. However, the approach discussed in that answer, which allows the number of satisfying combinations to be enumerated, provides nothing more than a preliminary step. You then have to collect the various combinations into groups, with the idea that each group of combinations would have the same scaling factor.
I emphasize that this entire approach is very crude. If you know generating functions, I suspect that is a far superior approach.
As another fine point:
Normally, when you are applying Stars and Bars theory to enumerate the number of non-negative integer solutions to
$$x_1 + x_2 + x_3 \leq n,$$
you would use the trick that
$$x_1 + x_2 + x_3 + c = n,$$
where the length of the word is $(n-c)$. Here, that approach is no good, because in (1) above, the numerator of the scaling factor represents the exact length of the word to be formed.
Therefore, I must take the (even more) crude approach of splitting the problem into $(7)$ cases, depending on the length of the word to be formed.
Further, for the smaller values of $n$, I won't even bother with the Stars and Bars analysis, since manual (brute force) enumeration will be easier.
In all of the analysis below, where I do employ Stars and Bars theory, the variables of $x_1, x_2, x_3$ will represent the number of a's, b's, and c's, respectively.
$\underline{\text{Case 1:} ~\text{Number of words with} ~1 ~\text{letter}}$
There are $3$ choices for the letter, so the number of possible words is
$$3.$$
$\underline{\text{Case 2:} ~\text{Number of words with} ~2 ~\text{letters}}$
There are $3$ possible double letter words, and $6$ possible words with two distinct letters.
The number of possible words is
$$3 + 6 = 9.$$
$\underline{\text{Case 3:} ~\text{Number of words with} ~3 ~\text{letters}}$
Since there are at least $3$ of each of the a's, b's, and c's, the number of possible combinations is the same as the number of non-zero integer solutions to
$$x_1 + x_2 + x_3 = 3.$$
In accordance with Stars and Bars, there are
$\binom{3+2}{2} = \binom{5}{2} = 10$ such combinations. Grouping them by their scaling factors:
$(3)$ triplets:
$~~\displaystyle 3 \times \frac{3!}{3! \times 0! \times 0!} = 3.$
$(6) = 3 \times 2$ - double + single:
$~~\displaystyle 6 \times \frac{3!}{2! \times 1! \times 0!} = 18.$
$(1)$ - 3 singletons:
$~~\displaystyle 1 \times \frac{3!}{1! \times 1! \times 1!} = 6.$
The number of possible words is
$$3 + 18 + 6 = 27.$$
$\underline{\text{Case 4:} ~\text{Number of words with} ~4 ~\text{letters}}$
Since the number of non-negative integer solutions to
$$x_1 + x_2 + x_3 = 4$$
is $\binom{4 + 2}{2} = \binom{6}{2} = 15$, I would ordinarily presume that $15$ satisfying combinations must be dealt with. Here, since there are only $3$ b's, the solution of
$$(x_1, x_2, x_3) = (0,4,0)$$
must be rejected. Therefore, there are $14$ satisfying combinations.
Grouping them by their scaling factors:
$(2)$ quadruplets:
$~~\displaystyle 2 \times \frac{4!}{4! \times 0! \times 0!} = 2.$
$(6) = 3 \times 2$ - triple + single:
$~~\displaystyle 6 \times \frac{4!}{3! \times 1! \times 0!} = 24.$
$(3)$ - 2 doubles:
$~~\displaystyle 3 \times \frac{4!}{2! \times 2! \times 0!} = 18.$
$(3)$ - 1 double, 2 singles:
$~~\displaystyle 3 \times \frac{4!}{2! \times 1! \times 1!} = 36.$
The number of possible words is
$$2 + 24 + 18 + 36 = 80.$$
$\underline{\text{Case 5:} ~\text{Number of words with} ~5 ~\text{letters}}$
The number of non-negative integer solutions to
$$x_1 + x_2 + x_3 = 5$$
is $\binom{5 + 2}{2} = \binom{7}{2} = 21$.
Following the strategy in the linked article, of those $21$ solutions, the number of them where the constraint that $x_2$ (i.e. the number of b's) is $\leq 3$ may be computed by computing the number of non-negative integer solutions to
$$x_1 + x_2 + x_3 = (5 - 4) = 1.$$
The number of such solutions is $\binom{1 + 2}{2} = \binom{3}{2} = 3.$ Therefore, the total number of satisfying combinations is $21 - 3 = 18.$
Grouping them by their scaling factors:
$(2)$ quintuplets:
$~~\displaystyle 2 \times \frac{5!}{5! \times 0! \times 0!} = 2.$
$(4) = 2 \times 2$ quadruplet + singleton:
$~~\displaystyle 4 \times \frac{5!}{4! \times 1! \times 0!} = 20.$
$(6) = 3 \times 2$ triplet + double:
$~~\displaystyle 6 \times \frac{5!}{3! \times 2! \times 0!} = 60.$
$(3)$ triplet + 2 singles:
$~~\displaystyle 3 \times \frac{5!}{3! \times 1! \times 1!} = 60.$
$(3)$ 2 doubles + 1 single:
$~~\displaystyle 2 \times \frac{5!}{2! \times 2! \times 1!} = 90.$
The number of possible words is
$$2 + 20 + 60 + 60 + 90 = 232.$$
$\underline{\text{Case 6:} ~\text{Number of words with} ~6 ~\text{letters}}$
Here, you have two constraint violations to consider:
- $x_1 \leq 5.$
- $x_2 \leq 3.$
I will let $T_0$ represent the total number of non-negative integer solutions, without any regard to the above constraints.
I will let $A_1$ represent the set of solutions where the constraint of $x_1 \leq 5$ is violated.
I will let $A_2$ represent the set of solutions where the constraint of $x_2 \leq 3$ is violated.
I will let $T_1$ represent the number of elements in $A_1$ added to the number of elements in $A_2$.
I will let $T_2$ represent the number of elements in $A_1 \cap A_2$.
Then, in accordance with Stars and Bars Theory, the number of satisfying combinations will be
$$T_0 - T_1 + T_2.$$
Note that since $6 + 4 > 6$, it is actually impossible for the constraints on $x_1$ and $x_2$ to be simultaneously violated. Therefore, you know that $A_1 \cap A_2$ is the empty set, and therefore $T_2 = 0.$
$\displaystyle T_0 = \binom{8}{2} = 28.$
$\displaystyle T_1 = \binom{0 + 2}{2} + \binom{2 + 2}{2} = 7.$
Therefore, the total number of satisfying combinations is $28 - 7 = 21.$
Grouping them by their scaling factors:
$(1)$ group of 6:
$~~\displaystyle 1 \times \frac{6!}{6! \times 0! \times 0!} = 1.$
$(4) = 2 \times 2$ groups of 5,1:
$~~\displaystyle 4 \times \frac{6!}{5! \times 1! \times 0!} = 24.$
$(4) = 2 \times 2$ groups of 4,2:
$~~\displaystyle 4 \times \frac{6!}{4! \times 2! \times 0!} = 60.$
$(2)$ groups of 4,1,1:
$~~\displaystyle 4 \times \frac{6!}{4! \times 1! \times 1!} = 60.$
$(3)$ groups of 3,3:
$~~\displaystyle 3 \times \frac{6!}{3! \times 3! \times 0!} = 60.$
$(6) = 3 \times 2$ groups of 3,2,1:
$~~\displaystyle 6 \times \frac{6!}{3! \times 2! \times 1!} = 360.$
$(1)$ groups of 2,2,2:
$~~\displaystyle 6 \times \frac{6!}{2! \times 2! \times 2!} = 90.$
The number of possible words is
$$1 + 24 + 60 + 60 + 60 + 360 + 90 = 655.$$
$\underline{\text{Case 7:} ~\text{Number of words with} ~7 ~\text{letters}}$
The analysis of the enumeration of the number of satisfying combinations is virtually identical with that of Case 6.
$\displaystyle T_0 = \binom{9}{2} = 36.$
$\displaystyle T_1 = \binom{3}{2} + \binom{5}{2} = 13.$
The number of satisfying combinations is
$\displaystyle T_0 - T_1 = 36 - 13 = 23.$
Grouping them by their scaling factors:
$(1)$ group of 7:
$~~\displaystyle 1 \times \frac{7!}{7! \times 0! \times 0!} = 1.$
$(2) = 1 \times 2$ groups of 6,1:
$~~\displaystyle 2 \times \frac{7!}{6! \times 1! \times 0!} = 14.$
$(4) = 2 \times 2$ groups of 5,2:
$~~\displaystyle 4 \times \frac{7!}{5! \times 2! \times 0!} = 84.$
$(2)$ groups of 5,1,1:
$~~\displaystyle 2 \times \frac{7!}{5! \times 1! \times 1!} = 84.$
$(4) = 2 \times 2$ groups of 4,3:
$~~\displaystyle 4 \times \frac{7!}{4! \times 3! \times 0!} = 140.$
$(4) = 2 \times 2$ groups of 4,2,1:
$~~\displaystyle 4 \times \frac{7!}{4! \times 2! \times 1!} = 420.$
$(3)$ groups of 3,3,1:
$~~\displaystyle 3 \times \frac{7!}{3! \times 3! \times 1!} = 420.$
$(3)$ groups of 3,2,2:
$~~\displaystyle 3 \times \frac{7!}{3! \times 2! \times 2!} = 630.$
The number of possible words is
$$1 + 14 + 84 + 84 + 140 + 420 + 420 + 630 = 1793.$$
Adding the total number of words possible, in each of the individual cases:
$$3 + 9 + 27 + 80 + 232 + 655 + 1793 = 2799.$$
