Number of 3 element subsets of ${1,2,...,n}$ with a certain property. Given the set $\{1,2,\dots,n\}$, I want to count the number of three-element subsets $\{x,y,z\}$ such that $|x-y|=1$ or $|x-z| = 1$ or $|y-z|=1$. I'm not familiar with how to deal with the "inclusive or" in a counting problem. 

As someone pointed out, we can equivalently count the number of triples $\{x,y,z\}$ such that $|x-y|&gt;1$ and $|x-z|&gt;1$ and $|y-z|&gt;1$.
You need a pair of successive integers and an arbitrary one. And then subtract those three integers that have two successive ones. That likes Inc-Exc Principle. You have $n-1$ choices for successive ones and then have $n-2$ choices for that arbitrary one. For subtracting, we know there successive integers have two successive ones that there is $n-2$ ones. So the answer is:
$$(n-1)(n-2) - (n-2) = (n-2)^2$$
I think your first have to built a two elements subset of two consecutive integers. There is $n - 1$ ways to do. Then, you chose the last element between the $n - 2$ elements that remain. Finally there is $(n - 1)(n - 2)$ such subsets.