Question

Let $b_{1} b_{2} b_{3} b_{4}$ be a $4$-element permutation with $b_{i} \in$ $\{1,2,3, \ldots \ldots ., 100\}$ for $1 \leq i \leq 4$ and $b_{i} \neq b_{j}$ for $i \neq j$, 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

$b _{ i } \in\{1,2,3 \ldots \ldots \ldots 100\}$
Let $A =$ set when $b _{1} b _{2} b _{3}$ are consecutive $n ( A )=\frac{97+97+\ldots \ldots+97}{98 \text { times }}=97 \times 98$
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 $=18915$