relationst

Question

How many relations are there which are reflexive antisymmetric and symmetric?

Comments

answer is 1

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@Ashwin

reflexive relation = 2^{(n^2 -n )}

symmetric = 2ⁿ  *  2^(n(n-1)/2)

asymmetric =  2ⁿ  *  3^(n(n-1)/2)

then

symmetric $\cap$ antisymmetric= 2ⁿ

Now 2^{(n^2 -n )} >2ⁿ

then reflexive $\cap$ 2ⁿ⁼2ⁿ

right?

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This is a correct explanation .