Number of relations on 6 element set which are both symmetric and reflexive but not anti-symmetric.
$begingroup$
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
asked Dec 15 '18 at 14:15
user3767495user3767495
3888
$endgroup$
add a comment |
$begingroup$
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
asked Dec 15 '18 at 14:15
user3767495user3767495
3888
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 15 '18 at 14:15
user3767495user3767495
3888
$endgroup$
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
asked Dec 15 '18 at 14:15
user3767495user3767495
3888
asked Dec 15 '18 at 14:15
user3767495user3767495
3888
asked Dec 15 '18 at 14:15
user3767495user3767495
3888
asked Dec 15 '18 at 14:15
user3767495user3767495
3888
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
add a comment |
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
add a comment |
1 Answer
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oldest
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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
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)
answered Dec 15 '18 at 15:08
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
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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
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)
answered Dec 15 '18 at 15:08
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
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
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)
answered Dec 15 '18 at 15:08
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
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
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)
answered Dec 15 '18 at 15:08
Sameer BahetiSameer Baheti
5168
$endgroup$
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
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)
answered Dec 15 '18 at 15:08
Sameer BahetiSameer Baheti
5168
answered Dec 15 '18 at 15:08
Sameer BahetiSameer Baheti
5168
answered Dec 15 '18 at 15:08
Sameer BahetiSameer Baheti
5168
answered Dec 15 '18 at 15:08
Sameer BahetiSameer Baheti
5168
5168
add a comment |
add a comment |
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$begingroup$
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