Let $b_{1} b_{2} b_{3} b_{4}$ be a 4 -element permutation with $b_{i} \in\{1,2,3, \ldots \ldots . .,

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 $b_{i} 
eq b_{j}$ for $i 
eq j$, 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

Professor PC Thomas Classes · Answer

(18915) :There are 98 sets of 3 consecutive integers and 97 sets of 4 consecutive integers [As all $b_{i}{ }^{\prime} s$ are different]\ $	herefore \quad$ Number of permutations of $b_{1} b_{2} b_{3} b_{4}$ $=$ number of permutation when $b_{1} b_{2} b_{3}$ are consecutive + number of permutations when $b_{2} b_{3} b_{4}$ are consecutive - number of permutations when $b_{1} b_{2} b_{3} b_{4}$ are consecutive $=97 	imes 98+97 	imes 98-97 \quad\left[\because b_{i} 
eq b_{j}, i 
eq jight]$ $=97 	imes 195=18915$\