The non-uniform probability of sums from the throw of multiple dice I'm reading The Drunkards Walk by Leonard Mlodinow. In the book, the author writes:

From a throw of three dice, a sum of 9 and 10 can be constructed in an equal combinations. However, the outcome (6, 3, 1) is more probable than the outcome (3, 3, 3) because there is just one way you can throw 3 threes, yet there are 6 ways you can throw a 6, 3 and a 1.

This is how I understand probability theory. However, the statement led me to ask the question: why is there is just one way to throw 3 threes? Despite the fact that the outcome is the same, there are a number of different ways to throw three 3s? That is, each of the three dice are distinct from the others, so why do we exclude the other combinations of rolling 3 threes?

Six ways to get $1,3,6$:
$$
\begin{array}{c|c|c|c|c|c}
\overbrace{
\begin{array}{rc}
\text{first die:} & 1 \\
\text{second die:} & 3 \\
\text{third die:} & 6
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 1 \\
\text{second die:} & 6 \\
\text{third die:} & 3
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 3 \\
\text{second die:} & 1 \\
\text{third die:} & 6
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 3 \\
\text{second die:} & 6 \\
\text{third die:} & 1
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 6 \\
\text{second die:} & 1 \\
\text{third die:} & 3
\end{array}} &
\overbrace{
\begin{array}{rc}
\text{first die:} & 6 \\
\text{second die:} & 3 \\
\text{third die:} & 1
\end{array}}
\end{array}
$$
One way to get $3,3,3$:
$$
\overbrace{\begin{array}{rc}
\text{first die:} & 3 \\
\text{second die:} & 3 \\
\text{third die:} & 3
\end{array}}
$$
What would be the other ways to get $3,3,3$ besides that one?
