Let's roll four dice. What is the probability that there is no "4" on any of the dice conditional on each dice having different values.
My answer is the following:
$$\frac{5 \times 4 \times 3 \times 2}{6 \times 5 \times 4 \times 3}$$
The denominator being all events with different values; The numerator being all events with different values but "4".
Is my reasoning correct?
Your reasoning seems fine.
You can also reason alternatively as follows:
Notice that since you are given that the four dice are all different, then your four rolls will include exactly $4$ of the $6$ values from $1$ through $6$. So any given value has a $\frac46$ chance of occuring.