Let ${\mathrm{b}}_1{\mathrm{b}}_2{\mathrm{b}}_3{\mathrm{b}}_4$ be a 4-element permutation with ${\mathrm{b}}_i\in \{1,2,3,\dots \dots ,100\}$ for $1\le i\le 4$ and ${\mathrm{b}}_i\ne {\mathrm{b}}_j$ for $i\ne j$, such that either $b_1,b_2,b_3$ are consecutive integers or ${\mathrm{b}}_2,{\mathrm{b}}_3,{\mathrm{b}}_4$ are consecutive integers. Then the number of such permutations $b_1{b}_2{b}_3{b}_4$ is equal to __________.

$$\begin{gathered} b_i\in \{1,2,3\dots \dots \dots 100\} \\ \text{ Let }A=\text{ set when }{b}_1{b}_2{b}_3\text{ are consecutive } \\ \mathrm{n}\left(\mathrm{A}\right)=\frac{97+97+\dots \dots +97}{98\text{ times }}=97\times 98 \\ \text{ Similarly when }{\mathrm{b}}_2{\mathrm{b}}_3{\mathrm{b}}_4\text{ are consecutive } \\ \begin{array}{l}n\left(B\right)=97\times 98\\ \mathrm{n}(\mathrm{A}\cap \mathrm{B})=\frac{97+97+------97}{98\text{ times }}=97\times 98\end{array} \end{gathered}$$

$\text{ Similarly when }{b}_2{b}_3{b}_4\text{ are consecutive }$

$$\begin{gathered} \begin{array}{l}\mathrm{n}\left(\mathrm{B}\right)=97\times 98\\ \mathrm{n}(\mathrm{A}\cap \mathrm{B})=97\\ \mathrm{n}\left(\mathrm{AUB}\right)=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)-\mathrm{n}(\mathrm{A}\cap \mathrm{B})\end{array} \\ \text{ Number of permutation }=18915 \end{gathered}$$