When calculating probability of a one pair in 5 card poker, why do we use $\binom{12}{3} \cdot \binom{4}{1}^3$ and not $\binom{48}{3}$?

Question

I know that when calculating the probability of getting one pair in 5 card poker, it is done like: $$ P(\text{one pair}) = \frac{\binom{13}{1} \cdot \binom{4}{2} \cdot \binom{12}{3} \cdot \binom{4}{1}^3}{\binom{52}{5}} $$

When accounting for the 3 other cards, why don't we use $\binom{48}{3}$ instead? I see that this would also count the times when there is a full house, and two pairs, however those alone do not make up the difference $\binom{48}{3}$ and $\binom{12}{3} \cdot \binom{4}{1}^3$. What hands am I missing/counting twice?

Answer 1

In your $\binom{13}1\binom42\binom{48}3$, where did you count the hand comprising two red kings, two black queens, and the jack of clubs?

You counted it twice. First, when you started with a pair of red kings, and caught the black queens among the other three cards. Again, when you started with a pair of black queens, and caugut the red kings among the other three cards.