Number of ways to choose an ordered pair  $\left(\mathrm{a},\mathrm{b}\right)$of numbers from the set $\left\{1,2,3,......,10\right\}$such that $\left|\mathrm{a}-\mathrm{b}\right|\le 5$ is 

$|\mathrm{a}-\mathrm{b}|\le 5$means $\left|\mathrm{a}-\mathrm{b}\right|=0$ or $1\le \left|\mathrm{a}-\mathrm{b}\right|\le 5$
$\left|\mathrm{a}-\mathrm{b}\right|=0\Rightarrow \mathrm{a}=\mathrm{b}$ which is possible in 10 ways 
Let $\mathrm{x},\mathrm{y},\mathrm{z}$ represent, numbers before, in between and after a and b , then
$\mathrm{x}+\mathrm{y}+\mathrm{z}=8. 0.\le \mathrm{y}\le 4$ and $\mathrm{x},\mathrm{z}\ge 0$
= coefficient of $x^8$ in ${\left(1+\mathrm{x}+..+{\mathrm{x}}^8\right)}^2\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4\right)=35$
Number of ways in which $\mathrm{a}\ne \mathrm{b}$ and is $=2\times 35=70.$( $\because$a and b can be in any order)
So, total number of ways .$=10+70=80$