Q82.Let b1b2b3b4 be a 4-element permutation with bi ∈{1, 2, 3, … … … , 100} for 1 ≤i ≤4 and bi ≠bj for i ≠j , such that either b1, b2, b3 are consecutive integers or b2, b3, b4 are consecutive integers. Then the number of such permutations b1b2b3b4 is equal to ______.
What This Question Tests
This problem involves counting permutations with a specific condition: either the first three or the last three elements are consecutive integers. It requires careful casework and handling overlaps using the Principle of Inclusion-Exclusion or by defining disjoint cases.
Concepts Tested
Formulas Used
nPr = n! / (n-r)!
Principle of Inclusion-Exclusion (if applicable, or careful casework)
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📋 Question Details
- Chapter
- Permutation & Combination
- Topic
- Permutations with conditions, Consecutive integers
- Year
- 2022
- Shift
- 29 Jun Shift 1
- Q Number
- Q82
- Type
- Numerical
- NCERT Ref
- Class 11 Mathematics Ch 7: Permutations and Combinations
More from this Chapter
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