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Please answer this:2+3 =

Original Message (ID '272048') By Bill Simpson:
This is stunningly slower than the previous iterative solution when given a list of 10^4 elements, but does show that Count can be forced to solve your problem. Count[Partition[numbers//.{x___,y_,y_,z___} -> {x,y,z}, 3, 1],{x_,y_,z_}/;(xz || x>y&&y