Number of relations on 6 element set which are both symmetric and reflexive but not anti-symmetric.












0












$begingroup$


I did this by imagining a Venn diagram.



Number of relations which are Reflexive and Symmetric would be given by



$2^{binom{n}{2}}$
.
Now, this also contains some Anti-symmetric relations.



Number of relations which are Reflexive, Symmetric and Anti-symmetric would be $2^n$



And hence final answer must be given by $2^{binom{n}{2}}-2^n$



Am I correct in my reasoning?






relations







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  • 1




    $begingroup$
    You should include the reasoning that leads you to these numbers. In particular, I don't see how you get $2^n$ for the number of reflexive, symmetric, and antisymmetric relations. It seems to me that there is only $1,$ but I could be wrong.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 14:42
















0












$begingroup$


I did this by imagining a Venn diagram.



Number of relations which are Reflexive and Symmetric would be given by



$2^{binom{n}{2}}$
.
Now, this also contains some Anti-symmetric relations.



Number of relations which are Reflexive, Symmetric and Anti-symmetric would be $2^n$



And hence final answer must be given by $2^{binom{n}{2}}-2^n$



Am I correct in my reasoning?






relations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You should include the reasoning that leads you to these numbers. In particular, I don't see how you get $2^n$ for the number of reflexive, symmetric, and antisymmetric relations. It seems to me that there is only $1,$ but I could be wrong.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 14:42














0












0








0





$begingroup$


I did this by imagining a Venn diagram.



Number of relations which are Reflexive and Symmetric would be given by



$2^{binom{n}{2}}$
.
Now, this also contains some Anti-symmetric relations.



Number of relations which are Reflexive, Symmetric and Anti-symmetric would be $2^n$



And hence final answer must be given by $2^{binom{n}{2}}-2^n$



Am I correct in my reasoning?






relations







share|cite|improve this question









$endgroup$




I did this by imagining a Venn diagram.



Number of relations which are Reflexive and Symmetric would be given by



$2^{binom{n}{2}}$
.
Now, this also contains some Anti-symmetric relations.



Number of relations which are Reflexive, Symmetric and Anti-symmetric would be $2^n$



And hence final answer must be given by $2^{binom{n}{2}}-2^n$



Am I correct in my reasoning?






relations




relations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 14:15









user3767495user3767495

3888




3888








  • 1




    $begingroup$
    You should include the reasoning that leads you to these numbers. In particular, I don't see how you get $2^n$ for the number of reflexive, symmetric, and antisymmetric relations. It seems to me that there is only $1,$ but I could be wrong.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 14:42














  • 1




    $begingroup$
    You should include the reasoning that leads you to these numbers. In particular, I don't see how you get $2^n$ for the number of reflexive, symmetric, and antisymmetric relations. It seems to me that there is only $1,$ but I could be wrong.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 14:42








1




1




$begingroup$
You should include the reasoning that leads you to these numbers. In particular, I don't see how you get $2^n$ for the number of reflexive, symmetric, and antisymmetric relations. It seems to me that there is only $1,$ but I could be wrong.
$endgroup$
– saulspatz
Dec 15 '18 at 14:42




$begingroup$
You should include the reasoning that leads you to these numbers. In particular, I don't see how you get $2^n$ for the number of reflexive, symmetric, and antisymmetric relations. It seems to me that there is only $1,$ but I could be wrong.
$endgroup$
– saulspatz
Dec 15 '18 at 14:42










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No, you are not. You should emphasize on the definition of Reflexive relations. The number of Reflexive, Symmetric, Anti-symmetric relations is only $1$ because we need to have all the diagonal elements of the grid



enter image description here



in our subset relations. $2^n$ $(n=6 text{here} )$ means that you are considering selections of the diagonal elements whether to take one or not. The answer should be $2^{15}-1$. (Possible selections of one side of non-diagonal elements with the other side corresponding non-diagonal element and all diagonal ones to be selected automatically "minus" case in which no non-diagonal elements are chosen which makes it Reflexive, Symmetric, Anti-symmetric altogether)






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    $begingroup$

    No, you are not. You should emphasize on the definition of Reflexive relations. The number of Reflexive, Symmetric, Anti-symmetric relations is only $1$ because we need to have all the diagonal elements of the grid



    enter image description here



    in our subset relations. $2^n$ $(n=6 text{here} )$ means that you are considering selections of the diagonal elements whether to take one or not. The answer should be $2^{15}-1$. (Possible selections of one side of non-diagonal elements with the other side corresponding non-diagonal element and all diagonal ones to be selected automatically "minus" case in which no non-diagonal elements are chosen which makes it Reflexive, Symmetric, Anti-symmetric altogether)






    share|cite|improve this answer









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      0












      $begingroup$

      No, you are not. You should emphasize on the definition of Reflexive relations. The number of Reflexive, Symmetric, Anti-symmetric relations is only $1$ because we need to have all the diagonal elements of the grid



      enter image description here



      in our subset relations. $2^n$ $(n=6 text{here} )$ means that you are considering selections of the diagonal elements whether to take one or not. The answer should be $2^{15}-1$. (Possible selections of one side of non-diagonal elements with the other side corresponding non-diagonal element and all diagonal ones to be selected automatically "minus" case in which no non-diagonal elements are chosen which makes it Reflexive, Symmetric, Anti-symmetric altogether)






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        No, you are not. You should emphasize on the definition of Reflexive relations. The number of Reflexive, Symmetric, Anti-symmetric relations is only $1$ because we need to have all the diagonal elements of the grid



        enter image description here



        in our subset relations. $2^n$ $(n=6 text{here} )$ means that you are considering selections of the diagonal elements whether to take one or not. The answer should be $2^{15}-1$. (Possible selections of one side of non-diagonal elements with the other side corresponding non-diagonal element and all diagonal ones to be selected automatically "minus" case in which no non-diagonal elements are chosen which makes it Reflexive, Symmetric, Anti-symmetric altogether)






        share|cite|improve this answer









        $endgroup$



        No, you are not. You should emphasize on the definition of Reflexive relations. The number of Reflexive, Symmetric, Anti-symmetric relations is only $1$ because we need to have all the diagonal elements of the grid



        enter image description here



        in our subset relations. $2^n$ $(n=6 text{here} )$ means that you are considering selections of the diagonal elements whether to take one or not. The answer should be $2^{15}-1$. (Possible selections of one side of non-diagonal elements with the other side corresponding non-diagonal element and all diagonal ones to be selected automatically "minus" case in which no non-diagonal elements are chosen which makes it Reflexive, Symmetric, Anti-symmetric altogether)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 15:08









        Sameer BahetiSameer Baheti

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