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Given set A has n elements. Find

1) Number of all reflexive relations?

2) Number of all symmetric relations?

3) Number of all reflexive and symmetric relations? Why is enumeration of transitive relations difficult?

1) Relation could be viewed as a digraph hence after the self loop edge for each vertices. There are ${n \choose 2}$ edges hence two choices for each i.e either a relation or not Hence answer:$ 2 ^ {n \choose 2} $

2) Here another choice of self loop. There are ${n \choose 2}$ + 1 edges hence two choices for each i.e either a relation or not Hence answer:$ 2 ^ {{n \choose 2}+1} $

Is this correct? Third one I am very confused how to approach.

Amrita
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    You can find the first question here and the second one here. The third one was shown as the top question among the related questions in the sidebar on the right. – Martin Sleziak Sep 03 '16 at 23:15
  • BTW it is generally advised to ask only one question per post, see meta. – Martin Sleziak Sep 03 '16 at 23:16
  • I only saw the first one, but the approach was using matrices which I am not very familiar with. I read somewhere that relations could be viewed as digraphs, so wanted to try it like this. Sorry about the multiple questions part. – Amrita Sep 04 '16 at 00:38
  • @Amita If you already saw that question but needed different solution or did not understood solution, you should have mentioned it in your post (and link to the question). Along the lines of the recommendations from this meta discussion. – Martin Sleziak Sep 04 '16 at 04:53
  • Sorry in the future I shall link any I have seen. Hopefully this approach adds something since for me at least approaching relation as a digraph is not obvious. – Amrita Sep 04 '16 at 18:10

1 Answers1

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1) When it comes to combinations, order doesn't matter, but in this case, the order of the two vertices picked does matter since we are working with a digraph. So instead of ${n \choose 2}$ possible edges, we have $2 { n \choose 2}$ possible edges and hence there are a total of $2^{2 {n \choose 2}}$ reflexive relations.

2) Since we are working with symmetric relations, now we can use ${n \choose 2}$ instead of $2{n \choose 2}$. For the self-loop, we don't have just one self-loop, we have $n$ self-loops each of which we have the choice of having or not. So we have a total of $2^{{n \choose 2}+n}$ symmetric relations.

3) This is the same as 2 except now we don't have to make any choices about self-loops so the answer is simply $2^{{n \choose 2}}$

benguin
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