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Let $X$ = {$a,b,c,d,e$}. Let us call a binary relations $R$ on $X$ special if it satisfies all of the following conditions: (i) $R$ is reflexive, (ii) $R$ is symmetric and (iii) $R$ contains the pair ($a,b$). Find the number of special binary relations on $X$ you need not simplify your answer.

What I want to know is that why the number of reflexive relation is not $1$? As I know only {$(a,a),(b,b),(c,c),(d,d),(e,e)$} is reflexive. So one.. I know it must be wrong. Is there anybody to let me know?

And also I think the relation must contain {{$(a,a),(b,b),(c,c),(d,d),(e,e),(a,b),(b,a)$} I do know know how to solve it from here.

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    The number of reflexive relations is not 1 because, any other relation that contains the set you have written out is also reflexive.2012-01-13
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    Have you in your class proved that the number of reflexive relations on a set of cardinality $n$ is $2^{n^2-n}$?2012-01-13
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    Such a relation is basically completely determined by a subset of the nine element set $\{(a,c),(a,d),(a,e),(b,c),(b,d),(b,e),(c,d),(c,e),(d,e)\}$. So the total should be $2^9$.2012-01-13

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