Problem Analysis #15

[THOUGHT] The set has 6 elements, so the full Cartesian product has 36 pairs. Reflexive relations must include the 6 diagonal pairs (a,a) through (f,f). With exactly 10 elements total, there are 4 additional off-diagonal pairs. Symmetry requires these to come in pairs: both (i,j) and (j,i) for i ≠ j. Thus, exactly 2 such unordered pairs {i,j} are needed. The number of possible unordered pairs {i,j} with i < j is \binom{6}{2} = 15. The number of ways to choose 2 of them is \binom{15}{2}. [/THOUGHT] [PYTHON] from math import comb print(comb(15, 2)) [/PYTHON]

[THOUGHT] The Python output confirms $\binom{15}{2} = 105$, which is the number of ways to choose 2 undirected pairs out of 15 possible, giving exactly 4 off-diagonal elements plus 6 reflexive ones for a total of 10 elements in a reflexive symmetric relation. [/THOUGHT] $\boxed{105}$

Explanation

The reasoning begins by noting that the set has 6 elements, requiring any reflexive relation to include the 6 diagonal pairs (e.g., (a,a), (b,b), up to (f,f)). With the relation having exactly 10 elements total, exactly 4 off-diagonal pairs must be present.

Symmetry mandates that off-diagonal pairs appear in symmetric pairs: for any (i,j) with i ≠ j, (j,i) must also be included. Thus, 4 off-diagonal pairs correspond to exactly 2 such symmetric pairs.

The total number of possible unordered pairs {i,j} with i < j from 6 elements is given by the combination formula $\binom{6}{2} = 15$.

The number of ways to select exactly 2 of these 15 unordered pairs is $\binom{15}{2}$, which equals 105, as verified by the Python output.

Each such selection, combined with the 6 reflexive pairs, forms a unique reflexive and symmetric relation with precisely 10 elements, confirming that the size of $\mathcal{S}$ is 105.