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(D) Let $n(A) = 10$. The total number of relations from $A$ to $A$ is $2^{n^2} = 2^{100}$.For a relation to be reflexive,all $10$ elements of the form

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$2^{90}-2^{45}$

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(D) Let $n(A) = 10$. The total number of relations from $A$ to $A$ is $2^{n^2} = 2^{100}$.For a relation to be reflexive,all $10$ elements of the form $(a, a)$ must be present in the relation.There are $100 - 10 = 90$ remaining elements in $A 	imes A$ that can either be present or absent.Thus,the total number of reflexive relations is $2^{90}$.For a relation to be symmetric,if $(a, b)$ is present,then $(b, a)$ must also be present for all $a 
eq b$.There are $\frac{10 	imes 9}{2} = 45$ such pairs of the form ${(a, b), (b, a)}$.For a reflexive and symmetric relation,the $10$ diagonal elements $(a, a)$ must be present,and for each of the $45$ pairs,we have $2$ choices (either both are present or both are absent).Thus,the number of reflexive and symmetric relations is $2^{45}$.The number of reflexive relations that are not symmetric is the total number of reflexive relations minus the number of reflexive and symmetric relations.Required number $= 2^{90} - 2^{45}$.

Let $A$ be a set consisting of $10$ elements. The number of non-empty relations from $A$ to $A$ that are reflexive but not symmetric is

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