TITLE: Number of sets $\{x,y\}\subset\{1,2,\dots,100\}$ such that $|x-y|>10$
QUESTION [8 upvotes]: Here's a question I ran in to recently:

How many pairs $\{x,y\}\subset\{1,2,\dots,100\}$ are there such that $|x-y|>10$.

I can solve this problem with addition: By assuming $x<y$, we have $89$ possibilities for $y$ when $x=1$, and $88$ possibilities for $y$ when $x=2$, and so forth. This leads to the solution
$$89+88+\dots+1=\frac{90\cdot89}{2}$$

However, since
  $$\dfrac{90\cdot 89}{2}=\binom{90}{2},$$
  it feels like there must be some nice combinatorial argument, but I've been failing to see it. Can someone enlighten me?

REPLY [3 votes]: Line up 100 dots are in a row, so any choice of two dots with at least 10 dots between them 
will give us such a pair. $\;\;$Let $d_1, d_2, d_3$ be the number of dots in the 3 gaps created, 
so $d_1+d_2+d_3=98$ with $d_2\ge10.\;\;$  If we let $e_1=d_1, e_3=d_3,$ and $e_2=d_2-10$,
we get the equation $e_1+e_2+e_3=88,\;\;$ and this has $\dbinom{90}{2}$ solutions in nonnegative integers.