Consider a 10-sided die that works as regular 6-side die in the sense that each of the 1 thru 10 is equally likely to occur. Roll three of these die and count all possible outcomes for for which the numbers are different.
Well, this gives $9 \cdot 8 \cdot 7 = 720 $ possible outcomes.
Now, pick three different numbers at random from the numbers $1,...,10$. In this case all posible outcomes are ${10 \choose 3} = 120$.
Now, the difference is because in the first case if we were to list the outcomes of the die $(x_1,x_2,x_3)$ this list is different from $(x_2,x_1,x_3)$ while in the second case choosing the numbers 1,2,3 is the same as choosing 2,1,3 so since we have 6 such permutation of three numbers, if we divide the 720 by $6$ we obtain the outcomes of the second case.
Is my reasoning correct? I am just trying to fully understand these basic counting techniques.
Your reasoning is incorrect.
Suppose we have a blue die, a green die, and a red die. Denote the number displayed by the blue die by $b$, the number displayed by the green die by $g$, and the number displayed by the red die by $r$. Then we can represent an outcome of rolling the three dice by the ordered triple $(b, g, r)$, where $b, g, r \in \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Since there are $10$ possible results for each of the three dice, there are $10^3 = 1000$ possible outcomes.
Suppose the numbers displayed by the three dice are all different. Then for each of the ten possible numbers displayed by the blue die, there are nine possible numbers that the green die could display since the number displayed by the green die cannot equal that of the blue die and eight possible outcomes for the red die since the number displayed by the red die cannot equal the numbers displayed by the blue and green dice. Hence, there are $10 \cdot 9 \cdot 8 = 720$ possible outcomes in which the numbers displayed by the three dice are all different.