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 S. Then the number of elements in S is ______.

$A = \{a,b,c,d,e,f\},\quad |A| = 6$

$R \subseteq A\times A:\quad R \text{ reflexive and symmetric}$

$|R| = 10$

$S = \{R : R \text{ reflexive, symmetric},\ |R|=10\}$

$\text{Find } |S|$

$1.\ \text{count all pairs; reflexive fixes the diagonal}$

$2.\ \text{symmetry} \Rightarrow \text{count the off-diagonal pairs}$

$3.\ |R|=10 \Rightarrow \text{count the choices}$

$|A\times A| = 6^2 = 36$

$\text{reflexive} \Rightarrow (a,a),(b,b),\ldots,(f,f)\in R\ \ (6\ \text{fixed})$

$\text{off-diagonal pairs} = 36 - 6 = 30$

$(x,y)\in R \Leftrightarrow (y,x)\in R \ \Rightarrow\ \binom{6}{2} = 15\ \text{pairs}$

$\text{choose } k \text{ of the } 15 \text{ pairs}$

$|R| = 6 + 2k$

$6 + 2k = 10 \Rightarrow 2k = 4 \Rightarrow k = 2$

$|S| = \binom{15}{2} = \frac{15\cdot 14}{2} = 105$

$\{(a,a),\ldots,(f,f),(a,b),(b,a),(c,d),(d,c)\}$

$6 + 2 + 2 = 10 \quad ✓$

$\boxed{|S| = 105}$