Step-by-step Solution:

1. Understand the given conditions:

   • The set is $A = \{a, b, c, d, e, f\}$. The size of the set is $n = |A| = 6$.
   • A relation $R$ on the set $A$ is a subset of the Cartesian product $A \times A$. The total number of possible ordered pairs in $A \times A$ is $n^2 = 6^2 = 36$.
   • The relation $R$ must satisfy three properties: a. Reflexive: For every element $x \in A$, the pair $(x, x)$ must be in $R$. b. Symmetric: If a pair $(x, y)$ is in $R$, then the pair $(y, x)$ must also be in $R$. c. Size: The relation $R$ must contain exactly 10 elements, i.e., $|R| = 10$.

2. Apply the reflexive property:

   • Since $R$ is reflexive, it must contain all pairs of the form $(x, x)$ for $x \in A$.
   • These pairs are: $(a, a), (b, b), (c, c), (d, d), (e, e), (f, f)$.
   • There are $n=6$ such pairs. These 6 elements are mandatorily in any relation $R$ that we are counting.

3. Apply the size property:

   • We are given that $|R| = 10$.
   • We have already accounted for 6 elements due to the reflexive property.
   • Therefore, we need to choose $10 - 6 = 4$ additional elements for the relation $R$.
   • These 4 elements must be pairs $(x, y)$ where $x \neq y$ (off-diagonal elements), because all diagonal elements are already included.

4. Apply the symmetric property:

   • The symmetric property states that if $(x, y) \in R$, then $(y, x) \in R$.
   • For the 4 off-diagonal elements we need to choose, they must come in pairs. If we choose $(x, y)$ where $x \neq y$, we must also choose $(y, x)$ to maintain symmetry.
   • This means the 4 additional elements must consist of 2 pairs of the form $\{(x, y), (y, x)\}$.
   • Let's say we choose the pairs $\{(x_1, y_1), (y_1, x_1)\}$ and $\{(x_2, y_2), (y_2, x_2)\}$ where $x_1 \neq y_1$, $x_2 \neq y_2$, and the unordered pairs $\{x_1, y_1\} \neq \{x_2, y_2\}$.

5. Reframe the problem as a combination problem:

   • Each symmetric pair of off-diagonal elements, $\{(x, y), (y, x)\}$, corresponds to a unique unordered pair of distinct elements $\{x, y\}$ from the set $A$.
   • Our task is to choose 2 such symmetric pairs. This is equivalent to choosing 2 distinct unordered pairs of elements from the set $A$.

6. Calculate the number of choices:

   • First, we need to find the total number of possible unordered pairs of two distinct elements from the set $A = \{a, b, c, d, e, f\}$. This is given by the combination formula $C(n, k)$ where $n=6$ and $k=2$. $\text{Number of available unordered pairs} = \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$ - There are 15 such unordered pairs (e.g., $\{a, b\}, \{a, c\}, \dots, \{e, f\}$), each corresponding to a symmetric pair of ordered pairs (e.g., $\{(a, b), (b, a)\}$).

7. Calculate the final answer:

   • From these 15 available symmetric pairs, we need to choose exactly 2 to form the 4 additional elements of our relation $R$.
   • The number of ways to choose 2 pairs from 15 is given by $\binom{15}{2}$. $\text{Number of ways} = \binom{15}{2} = \frac{15!}{2!(15-2)!} = \frac{15 \times 14}{2 \times 1} = 15 \times 7 = 105$ - Therefore, there are 105 possible relations $R$ that satisfy all the given conditions.

Conclusion:

The number of elements in the set $\mathcal{S}$ is 105.