Drawing without replacement: why is the order of draw irrelevant?

The probability of picking a red ball first and then a green ball is $$ \frac{6}{10} \cdot \frac{4}{9} $$ The probability of picking a green ball first and then a red ball is $$ \frac{4}{10} \cdot \frac{6}{9} $$ Notice that the numbers in the denominator are the same, while the numbers in the numerator are the same but in reverse order? Multiplication is commutative.

Another way of looking at this: we don't care about the process you go through in picking the balls, as long as it is fair: each possible outcome (i.e. each possible subset of $5$ of the $10$ balls, where we consider the balls as in principle distinguishable) has the same probability. If this is the case, you just need to count the number of outcomes that belong to the event you're considering, and divide by the total number of outcomes.

If you took into consideration the order, you would get the same result but the calculation would be a bit more difficult:

• all possible sequences of $5$ balls respecting order (as if they were distinguishable): $10\cdot 9\cdot 8\cdot 7 \cdot 6$
• all possible selections of $\color{red}{2}$ out of $\color{red}{6}$ red balls: $\color{red}{\binom{6}{2}}$
• all possible selections of $\color{green}{3}$ out of $\color{green}{4}$ green balls: $\color{green}{\binom{4}{3}}$
• all possible arrangements of the selected $\color{red}{2}+\color{green}{3}$ balls: $5!$

All together $$\frac{\color{red}{\binom{6}{2}}\cdot \color{green}{\binom{4}{3}} \cdot 5!}{10\cdot 9\cdot 8\cdot 7 \cdot 6} = \frac{\color{red}{\binom{6}{2}}\cdot \color{green}{\binom{4}{3}}}{\frac{10!}{5!\cdot 5!}}= \frac{5}{21}$$