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(C) For a relation $R$ on a set $A$ with $n$ elements to be reflexive,it must contain all $n$ diagonal elements $(x, x)$ for all $x \in A$. Here,the s

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$105$

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(C) For a relation $R$ on a set $A$ with $n$ elements to be reflexive,it must contain all $n$ diagonal elements $(x, x)$ for all $x \in A$. Here,the set is $\{a, b, c, d, e, f\}$,which has $n = 6$ elements. Thus,$R$ must contain the $6$ elements: $(a, a), (b, b), (c, c), (d, d), (e, e), (f, f)$.Since $R$ is symmetric,if $(x, y) \in R$,then $(y, x) \in R$. The remaining $n^2 - n = 36 - 6 = 30$ elements are off-diagonal. These $30$ elements form $15$ pairs of the form $\{(x, y), (y, x)\}$ where $x 
eq y$.We are given that $R$ contains exactly $10$ elements. Since $6$ diagonal elements are already included,we need to choose $10 - 6 = 4$ additional elements from the off-diagonal pairs. Because the relation must be symmetric,if we pick $(x, y)$,we must also pick $(y, x)$. Therefore,we must choose $2$ pairs from the $15$ available off-diagonal pairs.The number of ways to choose $2$ pairs from $15$ is given by the combination formula ${}^{15}C_2 = \frac{15 	imes 14}{2 	imes 1} = 105$.

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

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