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Question #74703

Difficulty: 8/10 | 🟢 Free

The Question

Let AA and BB be two distinct points in R3\mathbb{R}^3. For i=1,2,…,ni = 1, 2, \dots, n, let PiP_i and QiQ_i be the points dividing the line segment ABAB internally and externally in the ratio λi:1\lambda_i : 1 respectively, where λ1,…,λn\lambda_1, \dots, \lambda_n are distinct positive real numbers not equal to 1. Let S={P1,…,Pn,Q1,…,Qn}S = \{P_1, \dots, P_n, Q_1, \dots, Q_n\}. Let NkN_k be the number of bijections f:S→Sf: S \to S satisfying the following conditions: (I) ff maps every internal division point to an external division point, and every external division point to an internal division point. (II) Exactly kk internal division points map to their corresponding external division points (i.e., f(Pi)=Qif(P_i) = Q_i for exactly kk indices). (III) No external division point maps to its corresponding internal division point (i.e., f(Qi)≠Pif(Q_i) \neq P_i for all ii). If DmD_m denotes the number of derangements of mm distinct objects, which of the following represents the value of ∑k=0nNk\sum_{k=0}^{n} N_k?

Options

  • A.

    n!Dnn! D_n

  • B.

    (Dn)2(D_n)^2

  • C.

    (n!)2−n!Dn(n!)^2 - n! D_n

  • D.

    D2nD_{2n}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Section Formula Internals|Derangements (Inclusion-Exclusion)|Fundamental Principle of Counting (FPC)

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the application of the Fundamental Principle of Counting (FPC) alongside Derangements (Inclusion-Exclusion) in a partitioned set.

The geometric context of internal and external section formulas naturally partitions the 2n2n points into two sets of size nn.

The primary trap lies in failing to recognize that the bijection decomposes into two independent mappings (P→QP \to Q and Q→PQ \to P) that must be multiplied, not added.

Students often incorrectly calculate derangements for the total set (D2nD_{2n}), assume symmetry and calculate (Dn)2(D_n)^2, or fail to realize that summing the restricted permutations (nk)Dn−k\binom{n}{k} D_{n-k} over all kk simplifies elegantly to the total unconstrained permutations n!n! for that specific directional mapping.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterVector Algebra - Class 12
ConceptSection Formula Internals
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree