Let $b_1{b}_2{b}_3{b}_4{b}_5$ be a 5 element permutation with $b_i\in \{1,2,\mathrm{3.......100}\}$ for $1\le i\le 5$ and $b_i\ne b_j$ for $i\ne j$ such that either $b_1,b_2,b_3,b_4$ are consecutive integers or $b_2,b_3,b_4,b_5$ are consecutive integers. Then the number of such permutations $b_1{b}_2{b}_3{b}_4{b}_5$ is equal to

$A=b_1{b}_2{b}_3{b}_4$ are consecutive integers
{1, 2, 3, 4, $b_5$ } $\to$ 96 possibilities
{2, 3, 4, 5, $b_6$ } $\to$ 96 possibilities
$\begin{array}{l}.\\ .\\ .\\ .\\ .\end{array}$ 
{96, 97, 98, 99, $b_5$ } $\to$ 96 possibilities
{97, 98, 99, 100, $b_5$ } $\to$ 96 possibilities
                                      ______________
                                     $n\left(A\right)=96\times 97$
$B=b_2{b}_3{b}_4{b}_5$ are consecutive integers
{ $b_1$, 1, 2, 3, 4} $\to$ 96 possibilities
$\begin{array}{l}.\\ .\\ .\\ .\\ .\end{array}$ 
{$b_1,97,98,99,100$} $\to$ 96 possibilities
                                           _______________
                                          $n\left(B\right)=97\times 96possibilities$
$n(A\cap B)=96$ 
$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$ 
= $96\times 97+96\times 97-96$ 
= 18528