[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}