Sam rolls 3 dice one at a time and places them in a row. How many outcomes in which all three dice have different numbers are possible?

The first die rolled can be any number. The second can be any of 5 other numbers but what the first was. So 5/6

The third can be any of 4 other numbers, so 4/6.

There are 6^3=216 possible outcomes:

(5/6)(4/6)(6^{3})=120

6 posibilities on the first, 5 for the second, 4 for the third

6 x 5 x 4 = 120 possible different combinations

6 x 6 x 6 = 216 total possibilities including duplicate numbers

probability of different combinations would be 120 / 216

That would reduce down to 5 / 9

What are the total number of combinations? 6^3 = 216

What are the number of possibilities in which all dices have different numbers?

6P3 = 6 times 5 times 4 = 120

Probability = 120/216 = 5/9

Remember: desired possibility 3 5 1 is not the same as 5 3 1 or 1 5 3

Hence Permutations, not combinations.

The answer was given for permutations: 6! / 3! = 120

Combinations would yield 20.