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Question #41898

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{1, 2, 3, 4, 5\}$ . A relation $R$ on $S$ is a \textit{tournament} if for every pair of distinct elements $x, y \in S$ , exactly one of $xRy$ or $yRx$ is true, and $xRx$ is false for all $x \in S$ . For any $X \subseteq S$ , define $R_{odd}(X) = \{ y \in S \mid |\{ x \in X \mid xRy \}| \text{ is odd} \}$ . Define a binary operation $*$ on the power set $P(S)$ by: $A * B = A \Delta B \Delta R_{odd}(A \cap B)$ where $\Delta$ denotes the symmetric difference. Let $N$ be the number of tournaments $R$ on $S$ such that for every $A \in P(S)$ , there exists a unique $B \in P(S)$ satisfying $A * B = \emptyset$ . Find $N$ .

Options

A.
120
B.
1024
C.
3125
D.
32768

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

Invertibility in Binary Operations|Symmetry vs. Antisymmetry|Cardinality of Power Sets

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of abstract algebra (binary operations on power sets) and linear algebra over finite fields.

The trap lies in attempting to solve the invertibility condition purely through set theory or assuming that the cardinality of the power set ( $2^5$ ) directly dictates the number of valid relations.

By mapping the sets to indicator vectors in $\mathbb{Z}_2^5$ , the invertibility requirement transforms into the condition that every principal submatrix of $I+M$ must be non-singular.

Students who fail to recognize this isomorphism will be unable to deduce that the relation must be a transitive tournament (a DAG), effectively locking them out of the $n!$ combinatorial result.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Invertibility in Binary Operations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free