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

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Permutations and Combinations

Let

b_{1} b_{2} b_{3} b_{4}

be a 4-element permutation with

b_{i} \in {1, 2, 3, \dots\dots\dots, 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: 18915

Reading "consecutive integers" as a run

m, m + 1, m + 2

taken in increasing order, let

A

be the permutations in which

b_{1}, b_{2}, b_{3}

form such a run and

B

those in which

b_{2}, b_{3}, b_{4}

do. For

A

there are

98

possible runs (

m = 1, \dots, 98

) and

b_{4}

may be any of the remaining

97

numbers, so

n (A) = 98 \times 97 = 9506

; by the same reasoning

n (B) = 9506

. When both hold at once,

b_{1}, b_{2}, b_{3}, b_{4}

are four consecutive integers

m, m + 1, m + 2, m + 3

, of which there are

97

, so

n (A \cap B) = 97

. Hence

n (A \cup B) = 9506 + 9506 - 97 = 18915

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer