[THOUGHT] We need to count the number of relations R on the set {a,b,c,d,e,f} (size n=6) that are:

1. Reflexive: must contain all diagonal pairs (x,x) for x in the set.
2. Symmetric: if (x,y) in R then (y,x) in R.
3. Exactly 10 elements.

Since the relation is reflexive, it must contain the 6 diagonal elements. So we have 6 fixed elements. The remaining elements we can choose are from the off-diagonal pairs.

For a set of size n, the off-diagonal pairs come in unordered pairs { (x,y), (y,x) } with x≠y. There are C(n,2) such unordered pairs. For n=6, C(6,2)=15 unordered pairs. Each unordered pair corresponds to two ordered pairs (x,y) and (y,x). If we include one, symmetry forces the other.

So each unordered pair (i.e., each 2-element subset {x,y}) can be either:

• Not included in R (neither (x,y) nor (y,x) is in R)
• Included in R (both (x,y) and (y,x) are in R)

Thus, the total number of relations that are reflexive and symmetric is 2^{C(n,2)}.

But we need exactly 10 elements total in R. Since the diagonal gives 6 elements, we need exactly 4 more elements from off-diagonals.

Each unordered pair chosen contributes 2 elements (both ordered pairs). So if we choose k unordered pairs, we add 2k elements. We need 2k = 4 => k=2.

Thus we need to choose exactly 2 unordered pairs out of the 15 unordered pairs. The number of ways is C(15,2).

But wait: is that all? Let's verify: total elements = 6 + 2k = 6+4=10, correct.

So the number of such relations is C(15,2) = 105.

But we must check: Could there be any other way? Could we include only one ordered pair from an unordered pair? No, because symmetry forces both or none. So indeed only choice is which unordered pairs to include.

Thus answer is 105.

However, let's double-check with Python by enumerating all possible choices? For n=6, there are 2^{15}=32768 possible reflexive symmetric relations. We can iterate over all subsets of the 15 unordered pairs, compute total elements = 6 + 2*size(subset), and count those equal to 10. That will confirm.

We'll implement in Python. </think>

[PYTHON] import itertools

Set size

n = 6

Generate all unordered pairs (i,j) with i<j

unordered_pairs = [] for i in range(n): for j in range(i+1, n): unordered_pairs.append((i,j)) print("Number of unordered pairs:", len(unordered_pairs)) # should be C(6,2)=15

Count subsets of unordered_pairs of size k such that 6 + 2*k == 10

target_total = 10 count = 0 for subset in itertools.combinations(unordered_pairs, 2): # k=2 # total elements = 6 + 2*2 = 10 count += 1

print("Number of such relations:", count)

Alternatively, compute directly using math.comb

import math print("Direct combinatorial count:", math.comb(len(unordered_pairs), 2)) [/PYTHON]

[THOUGHT] The Python output shows that the number of such relations is 105. This is the final answer.

\boxed{105}