DISCLAIMER: this is a question already on the site, but I am seeking to gain clarity on a different approach to this problem.
Question is the probability of a full house (i.e., $3$ of a kind + $2$ of another kind) when drawing 5 cards at random from a standard deck of $52$ cards.
Solution 1) $$P(Full\ House) = \frac{13 \cdot 4C3 \cdot 12 \cdot 4C2}{52C5}$$
Solution 2) $$P(Full\ House) = \frac{52}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48}$$
Solution 1 provides the correct answer; however, solution 2 is incorrect by a factor of $10$. I believe I might need to introduce $5C2\ or\ 5C3 = 10$ into solution 2 to correct for the discrepancy, but I am unsure as to what the reasoning behind this might be.
What is the reason for the difference between solution 1 versus solution 2?
Your instinct is right about why the second answer is correct: it's missing a $_5 C_2$ (or equivalently, a $_5 C_3$). In the comments, you asked an insightful question: why should you multiply by that and not ($5!$)?
Running with the $5!$ instinct for a second, let's consider what it might represent: it would be the number of ways to rearrange all cards. So, for instance, your full house might be $K \diamondsuit, Q \diamondsuit, 5 \diamondsuit, 3 \clubsuit, 7 \clubsuit$. There are indeed $5!$ ways to rearrange those, such as:
$$3 \clubsuit, K \diamondsuit, 5 \diamondsuit, Q \diamondsuit, 7 \clubsuit$$ $$K \diamondsuit, 7 \clubsuit, Q \diamondsuit, 3 \clubsuit, 5 \diamondsuit$$ $$Q \diamondsuit, 5 \diamondsuit, K \diamondsuit, 7 \clubsuit, 3 \clubsuit$$
...but let's take a closer look at that last rearrangement. Notice that it has the same form as the "original" deal -- that is, $\diamondsuit, \diamondsuit, \diamondsuit, \clubsuit, \clubsuit$. That means we had already implicitly counted this case in the calculation $$\frac{52}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48},$$ so we've now overcounted it.
The picture is a little clearer now though, because we can think carefully about exactly what we've overcounted. Specifically, we've overcounted all the $\diamondsuit \diamondsuit \diamondsuit \clubsuit \clubsuit$ configurations. There are $3! \cdot 2!$ of these, corresponding to the $3!$ and $2!$ ways to rearrange the diamonds and clubs, respectively. So, the correct multiplier wasn't $5!$, but $\frac{5!}{3! \cdot 2!}$, or $_5 C_3 =$ $_5 C_2$.