Absolute Subtraction Mistake

We know that $|a-b| = a-b$ if and only if $a-b$ is non-negative, hence $a \geq b$ .

For $a > b$ , there are ${ 11 \choose 2 }$ pairs.
For $a = b$ , there are 11 pairs.
Hence, there are 66 pairs in total.

Note: Alternatively, we can list out the cases according to the value of $a$ , and get $1 + 2 + 3 + \ldots + 11 = 66$ .

We must have $a\geq b$ .

There are 121 possibilities, on which 11 are $a=b$ , so $121-11=110$ possibilities with as much where $a<b$ as where $a>b$ , so we divide 110 by 2 and get 55 possibilities where $a>b$ . Total is $11+55=\boxed{66}$

If we create a $11\times 11$ matrix such that its entry is $0$ if $row\: number>=column\: number$ and $1$ otherwise, then total number of $0$ 's in the matrix is $\frac{11^2-11}{2}+11=\boxed{66}$ .