Question 8

Let $A = \{a, b, c, d, e, f\}$. The number of elements in $A$ is $n = 6$.

1. A relation $R$ is reflexive if for every $x \in A$, $(x, x) \in R$. There are $6$ such elements: $(a, a), (b, b), (c, c), (d, d), (e, e), (f, f)$.
2. The total number of elements in $R$ is $10$. Since $R$ is reflexive, it must contain these $6$ diagonal elements.
3. The remaining $10 - 6 = 4$ elements must be chosen from the off-diagonal elements $(x, y)$ where $x \neq y$.
4. A relation $R$ is symmetric if $(x, y) \in R \implies (y, x) \in R$. Thus, off-diagonal elements must occur in pairs of the form $\{(x, y), (y, x)\}$.
5. To have $4$ additional elements, we must choose exactly $2$ such symmetric pairs.
6. The number of possible distinct pairs $\{x, y\}$ from $6$ elements is $\binom{6}{2} = \frac{6 \times 5}{2} = 15$.
7. We need to choose $2$ pairs out of these $15$ available pairs.
8. Total number of such relations = $\binom{15}{2} = \frac{15 \times 14}{2} = 105$. Final Answer: 105