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Question #40668

Difficulty: 8/10 | 🟢 Free

The Question

Let $x_1, x_2, \dots, x_6$ be positive real numbers satisfying $\sum_{i=1}^6 x_i = 12$ and $\sum_{i=1}^6 x_i^2 = 24$ . Let $S = \{1, 2, 3, 4, 5, 6\}$ . For each $i \in S$ , define the set $A_i = \{ i, (i + \lfloor x_i \rfloor) \pmod 6 \}$ , where $0 \pmod 6$ is taken as $6$ . Let $N$ be the number of permutations $\pi$ of $S$ such that $\pi(i) \notin A_i$ for all $i \in S$ . Find the value of $N$ .

Options

A.
80
B.
82
C.
96
D.
114

Check Your Answer

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

Arithmetic Mean Inequality|Inclusion-Exclusion Principle|Derangements

Why This Question Is Tricky

Common Mistakes

The problem tests the intersection of algebraic inequalities and advanced combinatorics.

Students must first use the RMS-AM (or Cauchy-Schwarz) inequality to deduce that all $x_i = 2$ .

The core trap lies in the subsequent derangement setup.

A student might hastily assume that avoiding $i$ and $i+2 \pmod 6$ is isomorphic to the standard Menage problem (avoiding $i$ and $i+1 \pmod 6$ ), which yields $M_6 = 80$ .

However, a shift of 2 on a set of 6 elements splits the forbidden bipartite graph into two disjoint cycles of length 6, whereas a shift of 1 forms a single cycle of length 12.

This structural change alters the rook polynomial, requiring a fresh Inclusion-Exclusion calculation that yields 82.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Arithmetic Mean Inequality
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free