1 votes 1 votes closed with the note: got the answer The Number of Relations, Which are both Reflexive and Symmetric but not Anti-Symmetric, on a set with 6 elements, are ____________? Set Theory & Algebra set-theory + – Satbir asked Dec 3, 2018 • closed Nov 1, 2019 by Satbir Satbir 1.2k views comment Share Follow See all 4 Comments See all 4 4 Comments reply Soumya Tiwari commented Dec 3, 2018 reply Follow Share 32768? 0 votes 0 votes himgta commented Dec 3, 2018 reply Follow Share 2^15-1?? 0 votes 0 votes himgta commented Dec 3, 2018 reply Follow Share @Soumya Tiwari it should be 32767 ,you also have to subtract the case when only diagonal elements are present! 0 votes 0 votes Soumya Tiwari commented Dec 3, 2018 reply Follow Share Thanks got your point :) 0 votes 0 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes There is only 1 relation which is reflexive, symmetric and anti-symmetric Number of symmetric relations which are reflexive = $\large 2^{\frac{n(n-1)}{2}} = $$\large 2^{\frac{6(5)}{2}} = 32768$ Total number of symmetric relations which are reflexive but not anti-symmetric = $32768 - 1 = 32767$ Mk Utkarsh answered Dec 3, 2018 • selected Dec 4, 2018 by Satbir Mk Utkarsh comment Share Follow See all 3 Comments See all 3 3 Comments reply Balaji Jegan commented Dec 3, 2018 reply Follow Share There is only 1 relation which is reflexive, symmetric and anti-symmetric How? 0 votes 0 votes Satbir commented Dec 4, 2018 reply Follow Share See the examples below (using elements)={relations possible which is reflexive , symmetric and anti symmetric} {0,1} ={(0,0),(1,1)} {0,1,2}={(0,0),(1,1),(2,2)} {0,1,2,3}={(0,0),(1,1),(2,2),(3,3)} {0,1,2,3,...,n}={(0,0),(1,1),(2,2),(3,3),...,(n,n)} and so on. It is clear that the given relations are reflexive and if we remove any one ordered pair from RHS then the relation would no longer be reflexive. Also ,if we add more ordered pair then the relations will remain reflexive. for symmetric relation if (a,b) is present then (b,a) should also be present for anti symmetric relation if (a,b) is present then (b,a) should not be present but if there is an exception that ordered pair like (a,a),(b,b)....(n,n) could be present. So in the above example if we add any ordered pair like (a,b) (lets say, (1,3) in third example) then we have to add (b,a)(i.e. (3,1)) also to make it symmetric and this would violate the anti symmetric property. on the other hand if we add any ordered pair like (a,b) (lets say, (1,3) in third example) then we do not add (b,a)(i.e. (3,1)) the relation would no longer be symmetric but would still satisfy the anti symmetric property. 0 votes 0 votes smsubham commented Dec 26, 2019 reply Follow Share @Balaji Jegan here's how https://math.stackexchange.com/questions/2164978/how-many-reflexive-symmetric-and-antisymmetric-relations-are-there-on-an-n-ele See bof comment. 0 votes 0 votes Please log in or register to add a comment.