Question

Consider a set \( S = \{a, b, c, d\} \). Then the number of reflexive as well as symmetric relations from \( S \to S \) is:

• A. \(1024\)
• B. \(256\)
• C. \(16\)
• D. \(64\)

Hint:

For relations on a set of \( n \) elements: \begin{itemize} \item Reflexive relations require all \( (a,a) \) to be included \item Symmetric relations treat \( (a,b) \) and \( (b,a) \) as a single choice \item Number of symmetric choices \( = 2^{\frac{n(n-1)}{2}} \) \end{itemize}

Answer 1

The Correct Option is D

To find the number of reflexive as well as symmetric relations on a set \( S = \{a, b, c, d\} \), we need to understand a few key concepts about relations, reflexivity, and symmetry.

A relation \( R \) on a set \( S \) is reflexive if, for every element \( x \in S \), the pair \((x, x)\) is in \( R \). Given the set \( S = \{a, b, c, d\} \), reflexivity requires that \((a, a), (b, b), (c, c),\) and \((d, d)\) are in \( R \). Four pairs for reflexivity are fixed.

A relation is symmetric if, whenever \((x, y)\) is in \( R \), \((y, x)\) is also in \( R \). For any pair of distinct elements, we have two options: either both pairs \((x, y)\) and \((y, x)\) are in the relation, or neither of them is.

We need to count how many such relations are possible: 

1. There are 4 pairs of the form \((x, x)\) which must be included in every reflexive relation. Thus, these don't contribute to variability.
2. For the pairs of the form \((x, y)\) where \( x \neq y \), we have total pairs:
   • Possible pairs without repetition are \((a, b), (a, c), (a, d), (b, c), (b, d), (c, d)\).
   • Each pair can either have both \((x, y)\) and \((y, x)\) present or absent. Thus, each pair contributes 2 choices:
   • There are 6 pairs, thus total symmetric pair choices: \(2^6 = 64\).

The answer is the total number of possible reflexive and symmetric relations, which is \(64\).

Therefore, the correct answer is 64.