Is the problem: 

There is a list of unique numbers and you permute the list and that a "local maximum" means that the maximum has adjacent smaller numbers on _both_ sides, then I think I understand: 1,2,3 has no max, while 1,3,2 does. Then one must keep the max from either end, so there are 

n! - 2 x (n-1)! such lists. No? 

Ah, now I see! The above is wrong, just as 5 3 1 2 4 has no local maximum, yet 5 1 3 2 4 does. Once the maximum of the n numbers is constrained to an end, one must ask if another number becomes the (only) local maximum, and on and on.