I thought that this would be an easy problem, and it seems really obvious to me that the answer would be:

$$ \frac{\binom{13}{1}^4 \binom{48}{3}}{\binom{52}{7}} $$

but for some reason this comes out to be around $3.69$ which is not a valid probability.

So then I thought of a different approach where instead of $\binom{48}{3}$ I used $48 * 47 * 46$ but this is obviously even larger.

I'm so confused why this approach doesn't work because I've done similar problems before in this manner. I'm assuming there has to be something that I am missing.

I then did some searching on here and came to an answer I'm confident in of $0.5696$ using Inclusion-Exclusion and counting the bad hands so my solution looked like:

$$ 1 - \frac{4\binom{39}{7} - 3!\binom{26}{7} + 4\binom{13}{7}}{\binom{52}{7}} = 0.5696...$$

I understand how the Inclusion-Exclusion Principle works but I'm wondering what events we're unioning here that it's taken into account. I'm also wondering if there is a more straightforward way of calculating the number of good hands where there is at least 1 card from each suit out of the 7 dealt.