Before You Begin

Question #40300

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{1, 2, 3, 4, 5, 6\}$ . A permutation $f: S \to S$ is called \textit{acceptable} if it satisfies at least one of the following two conditions:

(I) $f(i) \neq i$ for all $i \in \{1, 2, 3\}$ (II) $f(j) \neq j$ for all $j \in \{4, 5\}$

Find the number of acceptable permutations of $S$ .

Options

A.
621
B.
576
C.
411
D.
309

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Universal Set Relativity|Derangements (Inclusion-Exclusion)|Complementary Counting (Class 11)|Complementary Counting (Class 12)

Why This Question Is Tricky

Common Mistakes

The question tests the Principle of Inclusion-Exclusion (PIE) and the ability to correctly define the universal set when calculating the intersection of complementary events.

The primary trap lies in using De Morgan's laws to find the complement $|\overline{A} \cap \overline{B}|$ (where $\overline{A}$ means at least one of $\{1,2,3\}$ is fixed, and $\overline{B}$ means at least one of $\{4,5\}$ is fixed).

Students often incorrectly calculate this intersection by simply choosing one fixed point from each set and permuting the rest ( $\binom{3}{1} \times \binom{2}{1} \times 4! = 144$ ), leading to the trap answer of $720 - 144 = 576$ .

This flawed approach fails to account for the overlapping sub-cases (e.g., fixing elements 1, 4, and 5 simultaneously), which double-counts permutations in the reduced sample space.

The rigorous method requires applying PIE to find $|A|$ , $|B|$ , and $|A \cap B|$ independently.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sets - Class 11
Concept	Universal Set Relativity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free