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Question #40604

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{1, 2, 3, \dots, 20\}$ . Three sets $A, B,$ and $C$ are defined as follows: $A = \{x \in S \mid \text{sgn}(x^2 - 15x + 50) \le 0\}$ $B = \{x \in S \mid \text{sgn}(x^2 - 21x + 104) \le 0\}$ $C = \{x \in S \mid \text{sgn}(x^2 - 12x + 27) \le 0\}$ where $\text{sgn}(\cdot)$ denotes the signum function. Find the number of ways to select 3 distinct elements from $S$ such that the selection contains at least one element from $A \cap B$ and at least one element from $C$ .

Options

A.
80
B.
174
C.
324
D.
394

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

Restricted Selection (At least one common)|Cardinality Inclusion-Exclusion|Signum Function Discontinuity

Why This Question Is Tricky

Common Mistakes

The core trap lies in the definition of the sets using the signum function inequality $\text{sgn}(f(x)) \le 0$ .

Students frequently assume this strictly implies $f(x) < 0$ because they associate the signum function's negative output with negative inputs, completely forgetting the jump discontinuity at zero where $\text{sgn}(0) = 0$ .

This omission leads to excluding the boundary roots of the quadratic equations, causing an off-by-one error in the cardinality of sets $A$ , $B$ , and $C$ .

When these flawed sets are processed through the Principle of Inclusion-Exclusion for restricted combinations, the missing elements drastically alter the intersection sizes (e.g., $A \cap B$ drops from size 3 to 1), cascading into a highly plausible but incorrect final answer of 80 instead of 394.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Restricted Selection (At least one common)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free