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Question #40326

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{1, 2, 3, 4, 5, 6\}$ and $A = \{1, 2, 3\}$ . A function $f: A \to S$ is chosen at random. Given that $f(1) \neq f(3)$ and there exists at least one $x \in A$ such that $f(x) \neq x$ , what is the probability that $f(x) \neq x$ for all $x \in A$ ?

Options

A.
$\frac{105}{179}$
B.
$\frac{7}{12}$
C.
$\frac{104}{179}$
D.
$\frac{25}{36}$

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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

Cartesian Product Cardinality|Sample Space Partitioning|Conditional Probability Foundations|Complementary Counting

Why This Question Is Tricky

Common Mistakes

The question tests conditional probability within a Cartesian product space (functions from A to S).

The primary trap is the failure to subtract the identity case from the reduced sample space.

Students easily calculate the number of functions satisfying $f(1) \neq f(3)$ as 180, but often overlook that the identity function $f(x)=x$ is embedded within this set and must be excluded due to the second condition.

This leads to the classic 180 vs 179 denominator error.

A secondary trap involves complementary counting in the numerator, where students must correctly identify and subtract the overlapping cases $f(1)=f(3)$ from the total derangements.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Cartesian Product Cardinality
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free