Why are we using combination in probability question when the objects are identical? 
There are $3$ blue , $4$ red ,and $5$ green identical balls in an urn. We draw $3$ balls without replacement. What is the probability that all of these $3$ balls have different colors?

$\mathbf{\text{Please read until the end }}$
This is a basic probabilty question. I solved it such that $$3! \times \frac{3}{12} \times \frac{4}{11} \times \frac{5}{10}=\frac{3}{11}$$
However, there is also another approach such that $$\frac{\binom{3}{1}\binom{4}{1}\binom{5}{1}}{\binom{12}{3}}=\frac{3}{11}$$
$\mathbf{\text{My Question}:}$ We know that when we use combination , it is valid for selecting $r$ objects among $n$ $\color{red}{\text{distinct}}$ objects. However , as you see , the question says that the balls are identical among each types. Then , why did we use combination ? For example , i think that using generating functions for finding denominator is more sensible than using combination ,i.e , $\binom{12}{3}$.
When i ask it to someone , they said that "oh okey , but it works". Hence , i want a good explanation from you guys. Do not say me " we use combination because it gives us all possible arrangemets,i.e $3!$ " ,  because i think that it cannot be the reason to use combination. I think also that we betray the definition of combination by using it to select identical objects.
Then what is the reason behind seeing these identical objects like "distinct" ? What is the reason behind using combination to select identical objects ? Thanks in advance !!
 A: Saying that the objects are identical really means that they are indistinguishable, except that claim is not really true since they can always be distinguished by their positions.  We can always label apparently indistinguishable objects to make them distinct.  It is useful to treat the balls as distinct since this gives us a uniform distribution, meaning that is equally likely that we draw any particular set of three balls.  With a uniform distribution, we can simply divide the number of ways of selecting a particular selection of colors by the total number of ways of selecting three of the twelve balls.
For instance, there is only one way to grab all three blue balls (up to order of selection), while there are $\binom{4}{3} = 4$ ways to select three of the four red balls and $\binom{5}{3} = 10$ ways to select three of the five green balls.  If we treated all balls of the same color as indistinguishable, we would not be able to explain why there is a higher probability of selecting three green balls than there is of selecting three blue balls.
A: Distinct (i.e., $\ne$ each other) objects can be "identical" in the sense that they can be indistinguishable.
The reason for calling the balls "identical" here is just to suggest that any two sets of $k$ balls are equally likely to be drawn.
The balls are not what philosopher Aristotle called "numerically identical": they are not the same ball.
For instance, at the factory, maybe one ball was produced before another, or was produced by Alice instead of by Bob, even if we cannot tell that now by looking at the balls.
A: There is no such thing as "3 identical blue balls". The sheer fact that we know there are three of them (or seventeen, or two thousand and twenty-nine) means that we can distinguish them if we want to. In other words: If object A and object B are truly identical, i.e. A=B, then there is only one object. To say that there are three blue balls means that I have a way to count them, means that I have a way to call one of them the first ($B_1$), one the second ($B_2$), and one the third ($B_3$), if needed.
But questions are usually worded like that! So what do they actually mean? They mean that even though there are three different blue balls, for the final outcome (when we count possibilities in a sample space) we are meant to ignore if any given blue ball is either $B_1$ or $B_2$ or ... The same with blue or green balls. I.e.  we are meant to regard $B_2 R_4 G_1$ and $R_1 B_3 G_2$  as "the same" outcome, likewise $R_2 B_1 B_2$ and $R_3 B_2 B_3$ as "the same" outcome, they are (part of) the outcomes "one blue one green one red" and "two blue, one red", respectively.
And it just so happens that for such models, where we call (strictly speaking: wrongly) some of the objects "identical" (by which we actually mean, they are distinct, but we identify some of them in our counting: we "forget" the distinction between them), then there are neat shortcut formulas for doing that.
It might be a good exercise, sometimes, to do the calculations without "forgetting" the distinction. In a way you do that there. And it gives the same result, so all is good.

To illustrate that, let's look at the following much easier question: Say there are two blue balls and one red ball in an urn. We draw two balls. How many outcomes are there, and in how many of them, the drawn balls are of different colour?
Model/Method 0: We distinguish all the balls ($B_1, B_2, R$) and we also keep track of the order in which we drew them.
Sample space: $\{B_1 B_2, B_2 B_1, B_1 R, R B_1, B_2 R, R B_2\}$. Six outcomes which we could have computed e.g. via $3 \cdot 2$ or if you prefer, $\binom{3}{1} \cdot \binom{2}{1}$ (three possibilities on the first draw, two on the second).
Good outcomes: $\{B_1 R, R B_1, B_2 R, R B_2 \}$. Four possibilities, which we could have computed e.g. as $1 \cdot 2 + 1 \cdot 1 + 1\cdot 1$ or if we are super precise, $\binom{2}{1} + \binom{1}{1} + \binom{1}{1}$ (first ball red: both second draws are good; first ball $B_1$: need to draw $R$ now; first ball $B_2$: need to draw $R$ now.)
So the probability is computed as $\dfrac{\binom{2}{1} + \binom{1}{1} + \binom{1}{1}}{\binom{3}{1} \cdot \binom{2}{1}} = 4/6 =2/3$.
Model / Method 1: We still distinguish the balls, but "ignore" / "forget" the order in which we draw them.
Sample space $\{B_1 R, B_1 B_2, B_2 R \}$. Three possibilities, which we could have computed e.g. by going through method 0 and then halve the result (cumbersome), or with the shortcut $\binom{3}{2}$ which however, when you write it out, does exactly the same, it is just a nice shortcut notation (choosing two ("distinct") out of three objects, ignoring the order).
Good outcomes: $\{B_1 R, B_2 R \}$. Two of them, which we can compute as $\binom{2}{1} \cdot \binom{1}{1}$ (one of the balls can be either $B_1$ or $B_2$, the other has to be $R$).
So this method is the one which computes the probability as $\dfrac{\binom{2}{1}\binom{1}{1}}{\binom{3}{2}}=2/3$, and we keep $B_1$ versus $B_2$ "distinct".
Model / Method 2: We ignore the distinction between the two "identical" blues balls (and also the order in which we draw).
Sample space: $\{ B R, BB\}$. However, not all of these outcomes have the same probability. In fact, the good outcome $BR$ has probability $2/3$, while the other, $BB$, has probability $1/3$. We can get that from either of the above models. We can also get it through a different consideration like yours, which here would look like
$$ 2! \times \dfrac{2}{3} \times \dfrac{1}{3}. $$
But where do you get that from? It seems to go back to model 0, counting that one of the draws must be blue (which has probability $\frac{\#\{B_1, B_2\}}{\#\{B_1, B_2, B_3\}} = 2/3$), the other must be red (which has probability $\frac{\#\{R\}}{\#\{B_1, B_2, B_3\}} = 1/3$) and then multiplying that with the $2!$ possible orders in which such a pair of draws can occur.
So in a way, your method could be called method $1'$ in that it derives from method 0 just in a different way than method 1 does. If it seems "basic" to you that's good, but it only means you are more used to it. I personally would have chosen method 1 but that's only because I'm more used to it. Note that both methods somewhere still make a distinction between $B_1$ and $B_2$, which, to come back to the beginning, is equivalent to saying there are two blue balls, not just one of them.
