How many poker hands have exactly one pair?

Question

I know the answer is $C_{13}^4 \times C_4^1 \times C_4^2 \times C_4^1 \times C_4^1 \times C_4^1$. But what I did was $C_{13}^1 \times C_4^2 \times C_{48}^3$. I don't understand why we need to choose the suits for the remaining $3$ cards.

Answer 1

Your calculation does not avoid the mishap that the hand has two pairs or a full house.