Three dice are rolled. If the probability of getting different numbers on the three dice is $$\dfrac{p}{q}$$, where $$p$$ and $$q$$ are co-prime, then $$q - p$$ is equal to

We begin by noting that three dice are rolled, so the total number of outcomes is $$6^3 = 216$$.

Next, the favorable outcomes, where all three dice show different numbers, can be counted by considering that the first die has 6 choices, the second die has 5 choices, and the third die has 4 choices, which gives $$\text{Favorable} = 6 \times 5 \times 4 = 120$$.

Therefore, the probability can be expressed as $$\frac{p}{q} = \frac{120}{216} = \frac{5}{9}$$, and here $$p = 5$$ and $$q = 9$$ are coprime.

Subtracting these values yields $$q - p = 9 - 5 = 4$$.

Hence, the correct answer is 4.