Question

Let $b_1{b}_2{b}_3{b}_4$ be a $4$-element permutation with $b_i\in$ $\{1,2,3,\dots \dots .,100\}$ for $1\le i\le 4$ and $b_i\ne b_j$ for $i\ne j$, such that either $b_1,b_2,b_3$ are consecutive integers or $b_2,b_3,b_4$ are consecutive integers.
Then the number of such permutations $b_1{b}_2{b}_3{b}_4$ is equal to ______.

Answer 1

Answer: 18915

$b_i\in \{1,2,3\dots \dots \dots 100\}$
Let $A=$ set when $b_1{b}_2{b}_3$ are consecutive $n(A)=\frac{97+97+\dots \dots +97}{98\text{ times }}=97\times 98$
Similarly when $b_2{b}_3{b}_4$ are consecutive $N(A)=97\times 98$
$n(A\cap B)=\frac{97+97+------97}{98\text{ times }}=97\times 98$
Similarly when $b_2{b}_3{b}_4$ are consecutive $n(B)=97\times 98$
$n(A\cap B)=97$
$n(\Lambda UB)-n(\Lambda )+n(B)-n(\Lambda \cap B)$
Number of permutation $=18915$