Consider a set S = {a, b, c, d}. Then number of reflexive as well as symmetric relations from S → S are

To find the number of relations on a set $S$ that are both reflexive and symmetric, we can analyze the properties of such relations step-by-step.

Let the set $S=\{a,b,c,d\}$ have $n=4$ elements.

A relation $R$ on $S$ is a subset of the Cartesian product $S\times S$. The total number of elements in $S\times S$ is $n^2=4^2=16$.

Let's represent the elements of $S\times S$ in a grid or matrix form:
- Diagonal elements: $(a,a),(b,b),(c,c),(d,d)$. There are $n=4$ diagonal elements.
- Non-diagonal elements: There are $n^2-n=16-4=12$ non-diagonal elements. These elements form pairs of the form $\left\{\right(x,y),(y,x\left)\right\}$ where $x\ne y$. The number of such unordered pairs is $\frac{n^2-n}{2}=\frac{12}{2}=6$ pairs.

Now we apply the constraints of the relation $R$:
1. Reflexive Condition: For $R$ to be reflexive, all diagonal elements $(x,x)$ for every $x\in S$ must be included in $R$. Thus, there is only $1$ choice for each of the $4$ diagonal elements (they must be in $R$).
2. Symmetric Condition: For $R$ to be symmetric, if an ordered pair $(x,y)$ is in $R$, then $(y,x)$ must also be in $R$. This means that for each of the $6$ non-diagonal pairs $\left\{\right(x,y),(y,x\left)\right\}$, we have $2$ choices: either we include both $(x,y)$ and $(y,x)$ in $R$, or we exclude both of them from $R$.

Therefore, the total number of reflexive as well as symmetric relations is determined solely by the choices we make for the non-diagonal pairs:
$\text{Number of relations}=2^{\frac{n(n-1)}{2}}$

Substituting $n=4$:
$\text{Number of relations}=2^{\frac{4\times 3}{2}}=2^6=64$

The correct answer is 64.