Consider set A = {1, 2, 3}. The number of sym | vedclass

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(D) For a set $A$ with $n$ elements,the total number of symmetric relations is $2^{n(n+1)/2}$.Here,$n = 3$,so the total number of symmetric relations

Vedclass · Accepted Answer

$32$

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(D) For a set $A$ with $n$ elements,the total number of symmetric relations is $2^{n(n+1)/2}$.Here,$n = 3$,so the total number of symmetric relations is $2^{3(4)/2} = 2^6 = 64$.$A$ symmetric relation must contain $(a, b)$ if it contains $(b, a)$.The pairs $(1, 2)$ and $(2, 1)$ are already included in the relation.This leaves the remaining pairs to be chosen: $(1, 1), (2, 2), (3, 3)$ (which are $3$ diagonal elements) and the pairs ${(1, 3), (3, 1)}$ and ${(2, 3), (3, 2)}$ (which are $2$ pairs of off-diagonal elements).Each of the $3$ diagonal elements can either be present or absent ($2^3$ ways).Each of the $2$ pairs of off-diagonal elements can either be present or absent ($2^2$ ways).Total number of symmetric relations containing $(1, 2)$ and $(2, 1) = 2^3 	imes 2^2 = 8 	imes 4 = 32$.

Consider set $A = \{1, 2, 3\}$. The number of symmetric relations that can be defined on $A$ containing the ordered pairs $(1, 2)$ and $(2, 1)$ is:

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