Let b 1 b 2 b 3 b 4 be a 4 -element permutation with b i ∈ 1 , 2 , 3 , … … … , 100…

$b_i\in \left\{1,2,3\dots \dots \dots 100\right\}$

Let $A=$ set when $b_1{ b}_2{ b}_3$ are consecutive

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

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 $=9506+9506-97=18915$