The probability of rolling 5 dice ordered or unordered

We roll 5 dice that have 6 faces.

If we roll them simultaneously what is the probability that no two of them are the same.

$$\frac{\binom{n}{r}}{\binom{n+r-1}{r}}=\frac{\binom{6}{5}}{\binom{10}{5}}$$

What is the difference between the above:

and the following:

$$\frac{6\cdot 5\cdot 4\cdot 3 \cdot 2 }{6^5}$$

In the first one we consider that they are unordered and in the second one we consider them unordered.

By applying stars and bars, $ \displaystyle {5 + 6 - 1 \choose 5}$ is the number of ways in which numbers from $\{1, 2, 3, 4, 5, 6\}$ can show up on five identical dice. For example, there can be two $2'$s, three $3'$s or all $5'$s and so on.

In other words, $ \displaystyle {10 \choose 5}$ counts solutions to the number of $6$-tuples of non-negative integers whose sum is $5$:

$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 5 ~, ~$ where $x_i$ is non-negative integer.

But the outcomes in this sample space are not equally probable and cannot be applied to calculate probability

To calculate probability, we first choose any $5$ numbers by $ \displaystyle {6 \choose 5}$ and multiply by $5!$ as the chosen numbers can show up in different orders on five dice. Then we note that the probability of a particular number showing up on a die is $\dfrac 16$. So the answer is

$ \displaystyle {6 \choose 5} \cdot \frac{5!}{6^5} = \frac{5}{54}$

You can also write the probability as,

$ \displaystyle \frac{5}{6} \cdot \frac{4}{6} \cdot \frac{3}{6} \cdot \frac{2}{6} = \frac{5}{54}$

That is the first die can be any number then for the second die there are $5$ choices, for the third die there are $4$ choices and so on...