What is the probability of getting two spades in five draws? For easier viewing, here is the question:

What is the probability of getting two spades from five draws


I understand how to solve this question using permutations. I have explained my work below, but I need help understanding why an approach involving combinations also works! That question is below the second line. 
My approach: I reasoned that
$$\frac{13\mathbb{P}2 * 39\mathbb{P}3}{52\mathbb{P}5}$$
would give me all possible card hands for a set arrangement of events. That is, the value above would represent the probability of getting some card hand with, say, this arrangement:
{Drew a spade, Drew a spade, Didn't, Didn't, Didn't}.
To get the probability of having any one such arrangement, I needed to multiply the above quantity by $\frac{5!}{3!2!}$, which is the number of different arrangements of twice drawing a spade and thrice not.

Now, an acceptable approach is also to use combinations. As follows:
$$\frac{13\mathbb{C}2 * 39\mathbb{C}3}{52\mathbb{C}5}$$
That's it! I don't understand what logic underpins setting up the calculation above. I would reason, at first glance, that this approach shouldn't work because
1) It ignores permutations! One possible arrangement is Spade(K), Spade(Q), Club (5), Club (4), Club(3). While a combination might take care of this possibility, it will not consider Spade(Q), Spade(K), Club (5), Club (4), Club(3) as being any different.
2) It ignores the distribution of events. $AAA^{c}A^{c}A^{c}$ is not considered distinct from $AA^{c}AA^{c}A^{c}$
 A: Despite the differences in nuances about order of cards being (or not being) relevant, keep in mind what you are asked to find the probability of makes absolutely no mention of order of cards.
When approaching via permutations, you consider order to be important and you let your sample space be $\underline{\text{all ways of drawing five cards in a specific order}}$.  As such, you count how many outcomes from the sample space will satisfy the requirement of having exactly two spades.
Notice that every outcome in our sample space is equally likely to occur.  This is what is most important in problems like these and lets us use the powerful result that the probability is the number of favorable outcomes divided by the total number of outcomes in the sample space.
This yields the calculation: $\frac{5!}{2!3!}\cdot \frac{13\cdot 12\cdot 39\cdot 38\cdot 37}{52\cdot 51\cdot 50\cdot 49\cdot 48}$

As an alternative (and my preferred method, due to simpler notation) one may consider a different sample space.  Namely, the sample space is $\underline{\text{all ways of drawing five cards ignoring order}}$.  There is nothing intrinsically more or less correct about using this sample space compared to the other.  Note, however, that the outcomes in this sample space order doesn't matter and they are not technically the same as outcomes in the previous sample space.
Despite this, we again have the very nice result that every outcome in this sample space is equally likely to occur.  This again allows us to calculate the probability as the number of favorable outcomes divided by the total number of outcomes.
This yields the calculation: $\dfrac{\binom{13}{2}\binom{39}{3}}{\binom{52}{5}}$
After some algebraic manipulation, you will find that these two numerical results are exactly the same.  Which you use is up to you.  Just keep in mind that when using one where order matters that you remember to count everything and not leave any out, while counting the one where order doesn't matter to not accidentally overcount anything.

For a silly example to hopefully convince you further, consider the following experiment.  You have a bag of fair six-sided dice of various colors.  Say, for example, that there are four dice, being red, blue, green, and orange.  If you pick a die at random and throw it, what is the probability that it lands on a six?
You may approach using different sample spaces.  Perhaps you prefer the sample space where color and number both matter: $A=\{\color{red}{1},\color{red}{2},\color{red}{3},\color{red}{4},\color{red}{5},\color{red}{6},\color{blue}{1},\color{blue}{2},\dots,\color{orange}{6}\}$
We have $|A|=24$.  We note that the outcomes in $A$ are all equally likely to occur.  We count the number of favorable outcomes, being $|\{\color{red}{6},\color{blue}{6},\color{green}{6},\color{orange}{6}\}|=4$, and so the probability is $\frac{4}{24}$
Compare this to the sample space where color is unimportant: $B=\{1,2,3,4,5,6\}$.  We have $|B|=6$.  We note that the outcomes in $B$ are all equally likely to occur.  We count the number of favorable outcomes, being $|\{6\}|=1$, and so the probability is $\frac{1}{6}$.
In the second option, we don't really lose anything by ignoring the color on the die for this particular question since our question did not ask anything about the color.  In the same way, the card counting problems didn't matter which option you use.  Only when it makes mention of order being relevant for some reason (for example, what is the probability that being dealt a 5-card hand, the first card dealt is a spade and you receive exactly two spades overall) should you be forced to choose the first option.
