Given that three dice are thrown.

The total number of outcomes when three dice are thrown together is $6^3=6\times 6\times 6$.

The number of outcomes such that all the outcomes are different is $=6\times 5\times 4$.

For ex: If the outcome in dice $1$ is ${}^{"}{6}^{"}$ then the number of outcomes for dice $2$ should be $5 (1,2,3,4,5)$ which excludes the outcome ${}^{"}{6}^{"}$.

Hence, the required probability is $=\frac{6\times 5\times 4}{6\times 6\times 6}$

$=\frac{20}{36}=\frac{5}{9}=\frac{p}{q}$

$\Rightarrow q-p=9-5=4$

Therefore, the required answer is $4$.