Probability of full house?

Question

I have seen there are other answers on this, but they all follow the pattern

1. first, choose which card you have $3$ of
2. then, choose $3$ of the $4$ available cards
3. then, the card you have $2$ of
4. then, $2$ of $4$ cards

and they get to

$$\binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{2}$$

for the numerator.

My question is: doesn't this imply some kind of ordering?

By following the same logic, I would think that for the denominator

1. pick any of $52$ cards
2. then, pick any of $51$ remaining cards

and so on, getting $52 \cdot 51 \cdot 50 \cdot 49 \cdot 48$. But instead, the correct value of the denominator is $\binom{52}{5}$.

Why are those cases different?