Let the set of all relations $$R$$ on the set $$\{a, b, c, d, e, f\}$$, such that $$R$$ is reflexive and symmetric, and $$R$$ contains exactly 10 elements, be denoted by $$\mathcal{S}$$.

Then the number of elements in $$\mathcal{S}$$ is ________.

The underlying set is $$A=\{a,b,c,d,e,f\}$$ with $$|A|=6$$.

A relation on $$A$$ is a subset of $$A\times A$$. The required relations must satisfy two conditions:

1. Reflexive: for every $$x\in A$$, the ordered pair $$(x,x)$$ is in the relation.   Therefore the six diagonal pairs   $$\{(a,a),(b,b),(c,c),(d,d),(e,e),(f,f)\}$$   are compulsory.   So each admissible relation already contains $$6$$ pairs.

2. Symmetric: if $$(x,y)$$ is in the relation, then so is $$(y,x)$$.

The problem states that the relation must contain exactly $$10$$ ordered pairs. Since $$6$$ pairs are fixed by reflexivity, we still have to add $$10-6=4$$ more ordered pairs.

Because of symmetry, the non-diagonal pairs must be chosen in symmetric blocks: adding $$(x,y)$$ automatically forces us to add $$(y,x)$$. Consequently the extra $$4$$ pairs must come in $$\frac{4}{2}=2$$ symmetric blocks.

How many off-diagonal symmetric blocks exist? Each block corresponds to an unordered pair $$\{x,y\}$$ with $$x\neq y$$. The number of such unordered pairs in a 6-element set is $$\binom{6}{2}=15$$.

To create the relation we simply pick any $$2$$ of these $$15$$ unordered pairs; each chosen pair contributes its two ordered pairs and thus supplies the needed $$4$$ additional elements.

Hence the number of admissible relations is $$\binom{15}{2}=105$$.

Therefore, the set $$\mathcal{S}$$ contains $$\mathbf{105}$$ relations.