$\textbf{The Problem:}$ Let $X_1,\dots,X_{100}$ be independent absolutely continuous random variables that all have the same marginal density function. Find the probability that $X_{20}$ is the largest and $X_{12}$ is the smallest among the $100$ numbers.
$\textbf{Thoughts and Concerns:}$ Since the random variables are iid they are also exchangeable, and since they are absolutely continuous and independent they are also jointly continuous. Since they have a joint probability density function, the probability that two or more of these random variables are equal is zero. Therefore we can always pick the largest and the smallest uniquely, so that $$\sum_{i,j}P(X_i\text{ is the largest and }X_j\text{ is the smallest})=1.$$ Since the random variables are exchangeable we have that $P(X_i\text{ is the largest and }X_j\text{ is the smallest})$ does not depend on $i$ or $j$. Therefore we have that $$\begin{align}1&=\sum_{i,j}P(X_i\text{ is the largest and }X_j\text{ is the smallest})\\&=\binom{100}{2}P(X_1\text{ is the largest and }X_{100}\text{ is the smallest}).\end{align}$$ It follows that $$P(X_{20}\text{ is the largest and }X_{12}\text{ is the smallest})=\frac{1}{50\cdot 99}.$$
Is my reasoning correct? Any comments and suggestions are most welcome and highly appreciated.
There's an error in the two-line equation near the end. In the first line, $1=\sum_{i,j}\dots$, the $i$ and $j$ have to be an ordered pair because interchanging them gives another term of the sum ("smallest" $\neq$ "largest"). But in the second line, you treat $\{i,j\}$ as an unordered pair to get the count $\binom{100}2$. So I think your answer is off by a factor $2$.