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Number of relations that are both symmetric and reflexive

Let $X$ be a set with $8$ elements.

How many binary relations on $X$ are either reflexive or symmetric or both?

show work. you need not simplify the answer.

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marked as duplicate by Andres Caicedo, Austin Mohr, t.b., Asaf Karagila, Zev Chonoles Jan 10 '12 at 16:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Funny -- "show work" also appears in this answer, but there the request goes in the other direction... –  joriki Jan 10 '12 at 7:46
Added homework tag. @Q123: you're likely to get more helpful answers if you share what you've done so far (or at least what your thoughts are about how you might do the problem). –  Clive Newstead Jan 10 '12 at 8:09
Here is a start. For symmetric, there are $56$ ordered pairs of distinct elements. For any such pair, you can say yes or no. So there are $2^{56}$ symmetric relations. A similar argument (but be careful about equality) will count the symmetric relations. Add. The reflexive and symmetric have been double-counted. –  André Nicolas Jan 10 '12 at 8:14
It is the number which are reflexive plus the number which are symmetric minus the number which are both (to avoid double counting) –  Henry Jan 10 '12 at 8:15

1 Answer 1

up vote 1 down vote accepted

In a set with 8 elements, a binary relation, $R$ can be thought of as a set of pairs of elements of the set for which that relation is true. That is, $(a,b) \in R \leftrightarrow aRb$ is true. As such, if there are 8 elements, then there 64 possible pairs (order matters). If the relation is reflexive, then this implies that $(a,a) \in R$ for all $a$. So we know that 8 of the pairs must be in the relation. This leaves 56 pairs which can either be in or not be in. So there are $2^{56}$ possible reflexive relations.

Similar combinatorics can be applied to find the answer for symmetric and both.

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You are saying symmetric when you mean reflexive. –  Stefan Geschke Jan 10 '12 at 8:47
Oh, yeah, you are completely correct. How dumb of me. Fixed in most recent edit. –  Keith Irwin Jan 10 '12 at 21:40

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