No of antisymmetric relation : 2^n * 3^((n^2 – n)/2) ————–(1)

no of antisymmetric rel which are reflexive as well is : 3^((n^2 – n)/2) ………(2)

To get the answer of 1st question do eq. 1 – eq. 2 where ‘n’ is no of element in the set.

no of antisymmetric rel which are symmetric as well is : 2^n ……………..(3)

To get the answer of 2nd question do eq. 1 – eq. 3 where ‘n’ is no of element in the set.

Could you please add this reasoning for second problem solution? (a,a) kind of elements are allowed in both antisymmetric and symmetric. but for ab, if (a,b) and (b,a) both are in the relation then it can not be antisymmetric and if only(a,b) is in relation then it can not be symmetric. So finally we can not have any element in the form of (a,b) where ab. So a relation, which is both antisymmetric and symmetric, can contain only reflexive elements( means (a,a) kind of elements)

yes Sir, above reasoning is exactly correct for the second problem solution.

No of antisymmetric relation : 2^n * 3^((n^2 – n)/2) ————–(1) no of antisymmetric rel which are reflexive as well is : 3^((n^2 – n)/2) ………(2) To get the answer of 1st question do eq. 1 – eq. 2 where ‘n’ is no of element in the set. no of antisymmetric rel which are symmetric as well is : 2^n ……………..(3) To get the answer of 2nd question do eq. 1 – eq. 3 where ‘n’ is no of element in the set.