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Question #41027

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{z \in \mathbb{C} : |z| \le \sqrt{5} \text{ and } \text{Re}(z), \text{Im}(z) \in \mathbb{Z}\}$ . Let $V = \{z \in S : \text{Re}(z) \cdot \text{Im}(z) = 0\}$ and $C = S \setminus V$ . If all elements of $S$ are arranged randomly in a row to form a sequence, what is the probability that at least two elements of $V$ are adjacent?

Options

A.
$1 - \frac{\binom{13}{9}}{\binom{21}{9}}$
B.
$1 - \frac{\binom{12}{9}}{\binom{21}{9}}$
C.
$\frac{\binom{13}{9}}{\binom{21}{9}}$
D.
$1 - \frac{\binom{14}{9}}{\binom{21}{9}}$

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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

Complementary Counting|String Method (Adjacency)|Geometrical Interpretation of Modulus

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to decouple set theory and combinatorics from complex analysis.

The primary trap lies in the sample space definition: students must correctly identify the lattice points (Gaussian integers) within the geometric region $|z| \le \sqrt{5}$ and partition them based on the algebraic condition $\text{Re}(z)\text{Im}(z) = 0$ .

A common mistake is miscounting the boundary points or the origin, leading to incorrect cardinalities for sets $V$ and $C$ .

Furthermore, students might attempt to calculate 'at least two adjacent' directly using the inclusion-exclusion principle (string method) instead of recognizing that complementary counting combined with the gap method provides a direct, elegant path to the solution.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Complementary Counting
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free