Consider a set \( S = \{a, b, c, d\} \). Then the number of reflexive as well as symmetric relations from \( S \to S \) is:
Show 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}
Step 1: The set \( S \) has \( n = 4 \) elements.
Step 2: For a relation to be reflexive, all diagonal elements
\[
(a,a), (b,b), (c,c), (d,d)
\]
must be included.
Hence, there is no choice for these 4 ordered pairs.
Step 3: For a relation to be symmetric, if \( (x,y) \) is included then \( (y,x) \) must also be included.
The number of unordered distinct pairs is:
\[
\frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6
\]
Step 4: For each unordered pair \( \{x,y\} \), we have two choices:
\begin{itemize}
\item include both \( (x,y) \) and \( (y,x) \)
\item include neither
\end{itemize}
Hence, total number of such relations:
\[
2^6 = 64
\]
