This got bumped by a troll post that was deleted, but it seems like it's worth answering anyhow. First, let me rewrite it: Fix a positive integer $n$, and let $[n] = \{1, \ldots, n\}$. Given a permutation $\sigma: [n] \to [n]$, we say that the permutation $f: [n] \to [n]$ is good for $\sigma$ if there exists some $k \in [n]$ such that $f(\sigma(1)) < f(\sigma(2)) < \ldots < f(\sigma(k)) > f(\sigma(k + 1)) > \ldots > f(\sigma(n))$. Let $S(\sigma)$ be the set of permutations of $[n]$ that are good for $\sigma$. a) Prove that $|S(\sigma)| = 2^{n - 1}$. b) For $n \geq 4$, prove that there exist permutations $\sigma$ and $\tau$ such that $S(\sigma) \cap S(\tau) = \varnothing$. Part (a) is straightforward: first we count permutations $w: [n] \to [n]$ such that there exists $k$, $w(1) < \ldots < w(k) > \ldots > w(n)$. Each such permutation corresponds to the subset $\{w(1), \ldots, w(k - 1)\}$ of $[n - 1]$, and in fact this correspondence is easily seen to be a bijection. Thus, the number of such permutations is $2^{n - 1}$. For each such permutation $w$, we there exists a permutation $f = w \circ \sigma^{-1}$ that is good for $\sigma$. Since the set of permutations of $[n]$ forms a group, these $f$s are all distinct, and every $f$ good for $\sigma$ must be of this form. Part (b) seems trickier; maybe someone else can come up with a nicer argument. Consider choosing a permutation $\sigma$ at random. Because of the nice group structure, a given permutation $w$ is in $S(\sigma)$ with probability exactly $\frac{|S(\sigma)|}{n!} = \frac{2^{n - 1}{n!}$. Thus, if we choose two permutations $\sigma$ and $\tau$ independently at random, the expected number of permutations in $S(\sigma) \cap S(\tau)$ is precisely $n! \cdot \left(\frac{2^{n - 1}}{n!}\right)^2 = \frac{2^{2n - 2}}{n!}$. For $n \geq 7$, this number is less than 1; if the expectation is less than 1 then there must be actual values less than 1, and so in particular there is some pair $\sigma, \tau$ where the two sets have empty intersection. For $4 \leq n \leq 6$, this leaves us to find a pair of permutations by hand. We may as well take $\sigma$ to be the identity permutation; for $n = 4$ take $\tau = 2143$, for $n = 5$ take $\tau = 21435$, and for $n = 6$ take $\tau = 214356$. Edit: okay, I see, the probabilistic argument is not really necessary at all: we can just show that the pair $12\cdots n$ and $2143 56\cdots n$ works for any $n \geq 4$. But the probabilistic argument shows something more, namely, having $S(\sigma)$ and $S(\tau)$ intersect is in fact an extremely rare coincidence.