Number of Reflexive and Symmetric Relations with Exactly Ten Elements

To determine the number of relations $R$ on the set $A = \{a, b, c, d, e, f\}$ that are both reflexive and symmetric, and contain exactly 10 elements, we proceed with the following steps:

1. Reflexive Property: The set $A$ contains $n = 6$ elements. For $R$ to be reflexive, it must contain all $6$ diagonal pairs: $(a, a), (b, b), (c, c), (d, d), (e, e), (f, f)$ Hence, $6$ elements of the relation are fixed.

2. Symmetric Property: The total number of elements required in $R$ is $10$. Since $6$ elements are already accounted for by the reflexivity condition, the remaining number of non-diagonal elements needed is: $10 - 6 = 4$ For $R$ to be symmetric, whenever a non-diagonal pair $(x, y) \in R$ (with $x \neq y$), the pair $(y, x)$ must also belong to $R$. Therefore, non-diagonal elements must be selected in unordered pairs of the form $\{(x, y), (y, x)\}$. Each such pair contributes $2$ elements to $R$.

3. Counting the Ways to Choose the Pairs: To add $4$ non-diagonal elements, we must choose: $\frac{4}{2} = 2 \text{ unordered pairs of distinct elements from } A$ The total number of possible unordered pairs of distinct elements from the $6$-element set $A$ is: $\binom{6}{2} = \frac{6 \times 5}{2} = 15$ The number of ways to select $2$ pairs from these $15$ available pairs is: $\binom{15}{2} = \frac{15 \times 14}{2} = 105$

Thus, the number of elements in $S$ is 105.