[THOUGHT] To solve the problem, we need to count the number of relations $R$ on the set $\{a, b, c, d, e, f\}$ that are reflexive, symmetric, and contain exactly 10 elements.

1. Reflexive Property: A reflexive relation must include all pairs of the form $(x, x)$. For a set with 6 elements, there are 6 such pairs.
2. Symmetric Property: If $(x, y) \in R$, then $(y, x)$ must also be in $R$. This means non-diagonal pairs must be chosen in symmetric pairs.
3. Total Elements: The relation $R$ must have exactly 10 elements. Since 6 elements are already used for reflexivity, the remaining 4 elements must come from symmetric pairs of non-diagonal elements. Each symmetric pair contributes 2 elements, so we need $4 / 2 = 2$ symmetric pairs.
4. Counting Symmetric Pairs: The number of ways to choose 2 unordered pairs from the $\binom{6}{2} = 15$ possible unordered pairs of distinct elements is given by $\binom{15}{2}$.

Thus, the number of such relations is $\binom{15}{2}$.

[/THOUGHT] [PYTHON] import math

Calculate the number of ways to choose 2 unordered pairs from 15

n = 15 k = 2 result = math.comb(n, k) print(result) [/PYTHON]

[THOUGHT] The Python output confirms that the number of ways to choose 2 unordered pairs from 15 is 105. This matches the reasoning: after accounting for the 6 reflexive pairs, we need exactly 2 symmetric pairs (4 elements) to reach a total of 10 elements in the relation. Thus, the number of elements in $\mathcal{S}$ is 105. [/THOUGHT] $\boxed{105}$