TITLE: Why does the number of permutations of $n-1$ adjacent transpositions where the outputs are different equal $2^{n-2}$?
QUESTION [6 upvotes]: Maybe I'm wrong, but I just noticed that the different permutations of $(1,2)(2,3)(3,4),\dots,(n-1,n)$ seem to be $2^{n-2}$ and I don't know why this is true. Can someone help if I'm right about this and explain a little bit?
e.g.: $n=4$, $(1,2)(3,4)(2,3) = (3,4)(1,2)(2,3)$ and $(2,3)(1,2)(3,4) = (2,3)(3,4)(1,2)$ but $(1,2)(2,3)(3,4)$ and $(3,4)(2,3)(1,2)$ gives unique outputs. So the number of different permutations is $2^{4-2}=4$, I checked it for 5, 6, 7 and gives the same pattern.

REPLY [14 votes]: These are the Coxeter elements of the symmetric group, and they correspond to orientations of the Type A Dynkin diagram, of which there are $2^{n-2}$.