[THOUGHT] Let me analyze this problem step by step.

We have a set $S = \{a, b, c, d, e, f\}$ with 6 elements.

A relation $R$ on $S$ is a subset of $S \times S$.

For $R$ to be reflexive, it must contain all pairs $(x, x)$ for $x \in S$. So $R$ must contain: $(a,a), (b,b), (c,c), (d,d), (e,e), (f,f)$ - that's 6 elements.

For $R$ to be symmetric, if $(x, y) \in R$, then $(y, x) \in R$.

Since $R$ has exactly 10 elements and 6 of them are the diagonal elements (for reflexivity), we need exactly 4 more elements.

The non-diagonal pairs come in symmetric pairs: $\{(x,y), (y,x)\}$ where $x \neq y$.

The number of such symmetric pairs is $\binom{6}{2} = 15$ (choosing 2 distinct elements from 6).

Each such pair contributes 2 elements to the relation (both $(x,y)$ and $(y,x)$).

So if we choose $k$ symmetric pairs, we add $2k$ elements to the relation.

We need $2k = 4$, so $k = 2$.

Therefore, we need to choose 2 symmetric pairs from the 15 available.

The number of ways to do this is $\binom{15}{2}$.

Let me verify this with Python. [/THOUGHT] [PYTHON] from itertools import combinations

Set S = {a, b, c, d, e, f} represented as {0, 1, 2, 3, 4, 5}

S = {0, 1, 2, 3, 4, 5}

Diagonal elements (for reflexivity)

diagonal = {(x, x) for x in S} print(f"Diagonal elements: {len(diagonal)}")

Non-diagonal symmetric pairs: {(x, y), (y, x)} where x < y

We represent each such pair by the tuple (x, y) where x < y

symmetric_pairs = [(x, y) for x in S for y in S if x < y] print(f"Number of symmetric pairs: {len(symmetric_pairs)}")

We need R to have exactly 10 elements

6 diagonal + 4 non-diagonal = 10

4 non-diagonal means 2 symmetric pairs (each contributes 2 elements)

Count relations with exactly 10 elements that are reflexive and symmetric

count = 0

We need to choose 2 symmetric pairs from 15

for pair_combo in combinations(symmetric_pairs, 2): # Build the relation R = set(diagonal) # Start with diagonal for (x, y) in pair_combo: R.add((x, y)) R.add((y, x))

# Verify it has exactly 10 elements
if len(R) == 10:
    count += 1

print(f"Number of valid relations: {count}")

Verify with formula

from math import comb formula_result = comb(15, 2) print(f"Formula result C(15,2): {formula_result}") [/PYTHON]

[THOUGHT] The Python code confirms my analysis. We need to choose 2 symmetric pairs from 15 available pairs, which gives us $\binom{15}{2} = 105$ valid relations. [/THOUGHT]

\boxed{105}