Throw 4 equal dice such that all of the faces show different numbers

Imagine one throws 4 equal dice at the same time. How many combinations with different faces (in which no face appears twice) are possible? For the first die one has 6 possible faces, for the second 5, for the third 4 and for the fourth 3.

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You have it if you think order matters. You multiply the number of choices at each stage. If you think 1234 is the same as 3142, you need to divide by the number of orders, here $4 \times 3 \times 2 \times 1=24$