<s>[INST] Drawing without replacement: why is the order of draw irrelevant? I am trying to wrap my head around this problem:

Daniel randomly chooses balls from the group of $6$ red and $4$ green. What is the probability that he picks $2$ red and $3$ green if balls are drawn without replacement.

What I remember from my college days that the probability is found by this formula:
$$P(A)=\frac{\binom{6}{2}\binom{4}{3}}{\binom{10}{5}}=\frac{5}{21}$$
Is this correct? I am trying to understand why this works. Wouldn't probability depend on the order of balls drawn as the number of balls is changing after each draw? I get how we obtain numerator and denominator, I just feel that the probability should be dependent on the order. For example, the probability to pick red first is $\frac{6}{10}$ so the probability for the second draw becomes $\frac{5}{9}$ for red and $\frac{4}{9}$ for green. But if the first picked ball is green, the probability for the second draw becomes $\frac{6}{9}$ for red and $\frac{3}{9}$ for green. What am I missing?
 [/INST] You can comprehend the calculation in a simpler way with smaller numbers.

Daniel randomly chooses balls from the group of $3$ red and $2$ green. What
  is the probability that he picks $2$ red and $2$ green if balls are drawn
  without replacement.

Indeed we have to regard the order. There are $\frac{4!}{2!\cdot 2!}=6$ ways to  draw 2 red and 2 green balls:
$$\color{green}g\color{green}g\color{red}r\color{red}r, \color{green}g\color{red}r\color{green}g\color{red}r, \color{green}g\color{red}r\color{red}r\color{green}g, \color{red}r\color{green}g\color{green}g\color{red}r, \color{red}r\color{green}g\color{red}r\color{green}g, \color{red}r\color{red}r\color{green}g\color{green}g$$
Each way has the same probability: $\frac{3}{5}\cdot \frac{2}{4}\cdot \frac{2}{3}\cdot \frac{1}{2} \quad (ggrr)$
Multiplying with 6 (ways) we get $6\cdot \frac{3}{5}\cdot \frac{2}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}=\frac{3}5=0.6 $
Using binomial coefficients we get $\frac{\binom{3}{2}\cdot \binom{2}{2}}{\binom{5}{4}}=\frac{3\cdot 1}{5}=\frac35=0.6$
And we get the same result.
</s>