Let A = {0, 3, 4, 6, 7, 8, 9, 10} and R be th | vedclass

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(B) The set $A$ is given as $A = \{0, 3, 4, 6, 7, 8, 9, 10\}$. The number of odd elements is $3$ $(\{3, 7, 9\})$ and the number of even elements is $5

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$19$

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(B) The set $A$ is given as $A = \{0, 3, 4, 6, 7, 8, 9, 10\}$. The number of odd elements is $3$ $(\{3, 7, 9\})$ and the number of even elements is $5$ $(\{0, 4, 6, 8, 10\})$.The relation $R$ contains pairs $(x, y)$ such that $x - y$ is an odd positive integer or $x - y = 2$.$1$. Pairs where $x - y$ is an odd positive integer: Since $x - y$ is odd,one must be odd and the other even. There are $3 	imes 5 = 15$ such pairs where $x > y$.$2$. Pairs where $x - y = 2$: These are $(6, 4), (8, 6), (10, 8), (9, 7)$. There are $4$ such pairs where $x > y$.Total pairs in $R$ with $x > y$ is $15 + 4 = 19$.For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x)$ must also be in $R$. Since all $19$ pairs currently satisfy $x > y$,their symmetric counterparts $(y, x)$ where $y Therefore,we must add $19$ elements to $R$ to make it symmetric.

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