[THOUGHT] To solve this problem, we need to count the number of reflexive, symmetric relations on a set of 6 elements (a, b, c, d, e, f) that contain exactly 10 elements.

1. Reflexive Relations: A reflexive relation on a set of size $n$ must include all $n$ pairs of the form $(x, x)$. Here, $n = 6$, so the relation must include $(a, a), (b, b), \ldots, (f, f)$.

2. Symmetric Relations: A symmetric relation must include both $(x, y)$ and $(y, x)$ for any $x \neq y$.

3. Total Elements: The relation must have exactly 10 elements. Since the reflexive pairs account for 6 elements, the remaining 4 elements must come from symmetric pairs $(x, y)$ and $(y, x)$. Each such pair contributes 2 elements to the relation.

4. Counting Symmetric Pairs: The number of possible symmetric pairs (excluding the diagonal) is the number of ways to choose 2 distinct elements from 6, which is $\binom{6}{2} = 15$. We need to choose 2 such pairs (since each pair contributes 2 elements, and we need 4 more elements).

5. Combination Calculation: The number of ways to choose 2 pairs from 15 is $\binom{15}{2} = 105$.

Thus, the number of such relations is 105. [/THOUGHT] [PYTHON] from math import comb

Number of symmetric pairs (excluding diagonal)

n = 6 symmetric_pairs = comb(n, 2)

Number of ways to choose 2 pairs to add 4 elements (since each pair adds 2 elements)

result = comb(symmetric_pairs, 2)

print(result) [/PYTHON]

[THOUGHT] The Python output confirms that the number of reflexive, symmetric relations on a 6-element set with exactly 10 elements is 105. [/THOUGHT] $\boxed{105}$