Let b _ 1 b _ 2 b _ 3 b _ 4 be a 4 -element permutation with b _ i...

${b}_{i} \in \left\{1, 2, 3 \ldots \ldots \ldots 100\right\}$

Let $A =$ set when ${b}_{1}{b}_{2}{b}_{3}$ are consecutive

$n\left(A\right) = 97 + 97 + \ldots \ldots + 97\left(\text{added}98\text{times}\right) = 97 \times 98 = 9506$

Similarly when ${b}_{2}{b}_{3}{b}_{4}$ are consecutive

$n\left(B\right) = 97 \times 98 = 9506$

Now when ${b}_{1}{b}_{2}{b}_{3}{b}_{4}$ are consecutive then $n\left(A \cap B\right) = 97$

$n\left(AUB\right) = n\left(A\right) + n\left(B\right) - n\left(A \cap B\right)$

Number of required permutation $= 9506 + 9506 - 97 = 18915$