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← JEE Main 2022 · 29 Jun · Shift 1
MathematicsInteger+4 / -1Question 30 of 90

JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 30

Let b1b2b3b4b_1b_2b_3b_4 be a 4-element permutation with bi∈{1,2,3,…,100}b_i\in\{1,2,3,\dots,100\} for 1≤i≤41\le i\le4 and bi≠bjb_i\ne b_j for i≠ji\ne j, such that either b1,b2,b3b_1,b_2,b_3 are consecutive integers or b2,b3,b4b_2,b_3,b_4 are consecutive integers. Then the number of such permutations b1b2b3b4b_1b_2b_3b_4 is equal to ______.

Answer and worked solution

Official answer: 18915

Inclusion-exclusion. b1b2b3b_1b_2b_3 consecutive (increasing): 98 runs, 97 choices for b4b_4: 9506. Same for b2b3b4b_2b_3b_4: 9506. Both: b1..b4b_1..b_4 consecutive, 97.

Total 9506+9506−97=189159506+9506-97=18915.

Answer: 18915