bi∈1,2,3………100
Let A= set when b1b2b3 are consecutive
n(A)=97+97+……+97(added98times)=97×98=9506
Similarly when b2b3b4 are consecutive
n(B)=97×98=9506
Now when b1b2b3b4 are consecutive then n(A∩B)=97
n(AUB)=n(A)+n(B)−n(A∩B)
Number of required permutation =9506+9506−97=18915