order affecting the probability

There are $9$ balls - $4$ red, $3$ white and $2$ green balls in a bag. Total $4$ balls are taken out of that bag without replacement with the condition that exactly $2$ are red and $2$ are non-red. IMO, probability should be:

$$(4/9)*(3/8)*(5/7)*(4/6)$$

If I had taken out non-red balls for the first two times, the probability would've been:

$$(5/9)*(4/8)*(4/7)*(3/6)$$

Finally we will have only one sets of balls and probability is same irrespective of the order in which they are drawn - so why do we need to consider order?

Beginners often get puzzled/err on such problems.

Here no specific order has been given, so we need to consider all possible orders if we use the direct multiplication route to compute probability, thus the correct answer using the direct multiplication method should be

$(4/9)*(3/8)*(5/7)*(4/6)\times \frac{4!}{2!2!}$

Alternatively, it can be more simply computed as $\binom42\binom52 /\binom94$