A simple counting problem The question is as follows:

Two balls are drawn at random form a box containing ten balls numbered $0$ to $9$. Let the random variable $X$ be the maximum of the two numbers drawn and let $Y$ be the total of the two numbers drawn. If sampling is done without replacement, determine $X$ and $Y$'s probability functions.

So, my argument was as follows: there are $x$ $2$-subsets of $\{0,...,9\}=[9]$ with $x$ as the maximum elements, and ${10 \choose 2}$ total subsets of $[9]$. So
$$
P(X = x) = \frac{x}{{10\choose 2}}
$$
The suggested answer gives
$$
P(X = x) = \frac{2x}{10^{(2)}}
$$
Now I recognize these two are the same algebraically, but what is the intuition for the $10^{(2)}$ and $2x$? My understanding is that $10^{(2)}$ counts the number of arrangements of $2$ distinct elements from $10$, where order matters. Why should the order of the balls matter in this case? Is there any justification for why $10^{(2)}$ should be favoured over $10\choose 2$, or is it just two equivalent ways of thinking about things?
 A: I assume $10^{(2)} = _{10}P_2$.
You take the number of pairs (each of which corresponding to a unique maximum in a pair) as the denominator.
The second solution is identical, but uses a different process. The author took the amount of permutations in general, and then corrected for overcounting. Using $10^{(2)}$ differentiates between $4,9$ and $9,4$. In fact, each pair is counted twice. To correct for each pair being counted twice in the denominator, we multiply by two, yielding $\frac{2x}{10^{(2)}}$
A: Like you said, it seems that the book presents the results in a way that suggests that order might matter. In this case, since we only care about the maximum value, the order doesn't matter. Perhaps the book presents it this way so that the reader doesn't have to think of whether the order matters or not.
For example, if in another exercise in the book, the variable of interest is the first draw minus the second draw, then the reader would be more inclined to think of it with the mindset that the order matters, without having to think too much about it.
