I made a calculation for a much simpler problem, of which I'm not sure if it is correct or not. It is the following: having a bag with 20 red (R) and 10 white balls (W), in how many ways can you drawn 3 balls, one by one so that drawning one red ball and then two whites (WBB) is different to say, drawning a white one, then a red, and lastly a white (WBW).
My sample space would be
$$S = \left\{BBB, \; BBW, \; BWB, \; BWW, \; WBB, \; WBW, \; WWB, \; WWW \right\}$$
And the total number of ways of drawning three balls, where the order matters, is
$$n = \left(10\right)_3 + 3 \times (10)_{2} (20)_1 + 3 \times (20)_{2} (10)_1 + (20)_3 = 24360$$
where
$$(k)_l = \frac{k!}{(k-l)!}$$
which accouts for drawning $l$ balls from a total of $k$ balls, taking into account that you do not put the ball you drawned back to the bag.
My question is, how could I generalize this to having $k$ types of balls, where there are $l_{k}$ balls of each $k$ type, from a total of $n$ balls, where $\sum_{k} l_{k} = n$ (for example, in the problem above, $n=30$, $k = ${red, white}, $l_{red} = 20$ and $l_{white} = 10$, and $\sum_{k}l_{k} = l_{red} + l_{white} = 20+10 = 30$).
If you are satisfied with my answer in the comment, I'd like to recopy it here for the question to become 'officially' answered.
"I'm not sure whether I understand your reasoning correctly, however you assume that the balls in one colour are pairwise distinguishable, that is that there are e.g. $10⋅9⋅8$ ways to draw $BBB$. In such reasoning we don't even need colours, notice that you can just go for $30$ distinguishable balls. Even your calculations are correct, as drawing a sequence of $3$ distinguishable balls from a set of $30$ balls can be done in $30⋅29⋅28=24360$ ways"