A selection of objects (with replacement) using combination formula  
Consider an example where a pack contains 4 blue, 2 red, and 3 black pens. If a pen is drawn at random from the pack, replaced, and the process repeated 2 more times, what is the probability of drawing 2 blue pens and 1 black pen?

Now I am aware that a every single draw is an independent event and so we can use the multiplication rule of probability.
$$\text{Pr(drawing a blue pen)} = \frac{4}{9}$$
$$\text{Pr(drawing a black pen)} = \frac{3}{9}$$
$$\text{Pr(drawing a blue pen on first draw AND drawing a blue pen on the second draw AND drawing a black pen on the third draw)} = \frac{4}{9}\cdot\frac{4}{9}\cdot\frac{3}{9}$$
2 blue pens and 1 black pen can be arranged in $\frac{3!}{2!}$ ways so the total probability is $3\cdot\frac{4}{9}\cdot\frac{4}{9}\cdot\frac{3}{9}$.
My question is whether this can be done using just the combination formula?
I learned that you can select $r$ objects out of $n$ with replacement using $${n+r-1 \choose r}.$$
So we can select 2 blue balls out of 4 in ${4+2-1 \choose 2}$ ways and 1 black ball out of 3 using ${3 \choose 1}$ ways. We can choose 3 balls out of a total of 9 (with replacement) in ${9+3-1 \choose 3}$ ways.
According to me the probability using the combination formula should be
$$\frac {{5 \choose 2}{3 \choose 1}}{11 \choose 3}.$$ But this is not the same as what I got using the multiplication rule.
I would be really thankful if someone could guide me where I went wrong.
 A: The issue is that you wrongfully assume that each of the $\binom{11}{3}$ combinations are equally likely. 
Getting the combination of $\text{black}_1,\text{black}_1, \text{black}_1$ has a probability of occurring with $\frac{1}{9^3}$ but getting the combination of, say, $\text{black}_1, \text{black}_1, \text{blue}_1$ is not the same, but rather $\frac{3}{9^3}$.
This same mistake of thinking each outcome is equally likely can also be wrongly applied to say that the probability of flipping one head and one tails in two coin flips is $\frac13$. Someone might say "well there are three ways to choose an outcome on two coin flips without order mattering: $HH, HT, TT$. Therefore the probability of one head and one tails is $1/3$"
Perhaps it is easier to see why that method will not work in this more trivial example
A: If we write the generating function that depicts the drawing, we get :
$$  \left( \frac 4 9 b + \frac 3 9 k + \frac 2 9 r \right)^3 $$
The coefficient of $b^2k$ is then
$${3 \choose 2,1,0} \cdot \frac 4 9  \cdot \frac 4 9  \cdot \frac 3 9 $$
We do need the multinomial coefficient and the specific probabilities.
