Jingling, your answer is obviously wrong. Wxc's proof is beautif
A subset of valid placements: first put the heaviest on the right, then put others in any order on either side.
The number of such placements is (n-1)! * 2^{n-1}, which is al ready much greater than your answer when n is large enough.
The number of such placements is (n-1)! * 2^{n-1}, which is al ready much greater than your answer when n is large enough.