Calculating probability related to maximum of random variables Let $X_1, X_2, \cdots, X_n$ be non-negative continuous iid random variables. The goal is to find the probability:
\begin{align*}
\Pr(\max_{k+1 \leq i \leq j } X_i < \max_{1 \leq i \leq k }X_i)
\end{align*}
and I solve it by the following method:
\begin{eqnarray}
\Pr(\max_{k+1 \leq i \leq j} X_i \leq \max_{1 \leq i \leq k} X_i)=\int_0^\infty \Pr(\max_{k+1 \leq i \leq j} X_i \leq \omega) G_{\max_{1 \leq i \leq k} X_i} (d\omega)=\cdots=&\frac{k}{j}.
\end{eqnarray}
I guess that this problem can be solved by permutations methods. I was wondering if anyone could explain it.

It's simpler than that.
Note that we don't care about the $X_i$ for which $j < i \leq n$, since they aren't part of the problem statement.  In what follows, we'll simply assume the indices are in $1, \dots, j$.
First, observe that either $\max_{k+1 \leq i \leq j } X_i$ or $\max_{1 \leq i \leq k }X_i$, whichever is larger, is $\max_i X_i$.  Restating the question taking this into account leads to:
Let $X_1, X_2, \cdots, X_j$ be non-negative continuous iid random variables. The goal is to find the probability that $\arg\max_i X_i \leq k$.
At this point, it should be clear how to proceed.  Since each of the $j$ observations is equally likely to be the maximum due to the continuity and the i.i.d. properties, the probability that the maximum is in the first $k$ observations is $k/j$.
