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Let $\mathrm{b}_{1} \mathrm{~b}_{2} \mathrm{~b}_{3} \mathrm{~b}_{4}$ be a 4-element permutation with $\mathrm{b}_{i} \in\{1,2,3, \ldots \ldots, 100\}$ for $1 \leq i \leq 4$ and $\mathrm{b}_{i} \neq \mathrm{b}_{j}$ for $i \neq j$ , such that either $\mathrm{b}_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}$ 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_____________.
Option: 1
18915

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

1. If $\mathrm{b_{1},b_{2},b_{3}}$ are consecutive, then

(i) $\mathrm{b_{1}}$ can take any value from $\mathrm{1,2,3,...,98}$

(ii) $\mathrm{b_{2}}$ and $\mathrm{b_{3}}$ are automatically fixed

(iii) $\mathrm{b_{4}}$ can take any of the remaining $\mathrm{97}$ values

$\mathrm{\therefore 98\times97=9506}$

2. If $\mathrm{b_{2},b_{3},b_{4}}$ are consecutive, then

(i) $\mathrm{b_{4}}$ can take any value from $\mathrm{3,4,5,...,100}$

(ii) $\mathrm{b_{2}}$ and $\mathrm{b_{3}}$ are automatically fixed

(iii) $\mathrm{b_{1}}$ can take any of the remaining $\mathrm{97}$ values

$\mathrm{\therefore 98\times97=9506}$

3. Cases common in 1. and 2. ( i.e, when $\mathrm{b_{1},b_{2},b_{3},b_{4}}$ are consecutive )

(i) $\mathrm{b_{1}}$ can take any value from $\mathrm{1,2,3,.....,97}$

(ii) $\mathrm{b_{2},b_{3},b_{4}}$ are automatically fixed

$\mathrm{\therefore \text { Total }=9506+9506-97}\\$

$\mathrm{=18915}$

Hence answer is $\mathrm{18915}$

Posted by

Divya Prakash Singh

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