Question

The number of reflexive relations on a set \( A \) of \( n \) elements is equal to:

• A. \( 2^{n^2} \)
• B. \( n^2 \)
• C. \( 2^{n(n-1)} \)
• D. \( n^2 - n \)

Hint:

For relation counting problems: Total relations on \( n \) elements = \( 2^{n^2} \) Reflexive relations = Fix diagonal pairs, vary rest Formula to remember: \( 2^{n^2-n} \)

Answer 1

The Correct Option is C

Concept:
A relation on a set \(A\) with \(n\) elements is any subset of \(A \times A\).

Total ordered pairs in \(A \times A = n^2\).
Total number of relations \(= 2^{n^2}\).

A relation is reflexive if every element is related to itself.
That means all diagonal pairs must be present: \[ (a_1,a_1), (a_2,a_2), \ldots, (a_n,a_n) \] So these \(n\) pairs are fixed and cannot be chosen freely.

The remaining pairs: \[ n^2 - n = n(n-1) \] can either be included or excluded independently.

Step 1: Fix the reflexive pairs.
Reflexivity forces inclusion of all \(n\) diagonal pairs.

Step 2: Count remaining free pairs.
\[ \text{Remaining pairs} = n^2 - n = n(n-1) \]
Each of these pairs has 2 choices (include or exclude).

\[ \text{Number of reflexive relations} = 2^{n(n-1)} \]