Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?
Solution: $$\frac{\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1}}{\binom{52}{5}} = 0.047539$$
Why doesn't ${13\choose 1}{4\choose 2}{12\choose 1}{4\choose 2}{11\choose 1}{4\choose 1}\over{52\choose 5}$ OR ${13\choose 3}{4\choose 2}{4\choose 2}{4\choose 1}\over{52\choose 5}$ work?
On
Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:
$$ \binom{13}{2} \cdot \binom{4}{2} \cdot \binom{4}{2} \cdot \binom{13 - 2}{1} \cdot \binom{4}{1} = 123\,552 $$
Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in
In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.
In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.