# How to find the number of binary relations? [duplicate]

Possible Duplicate:
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.

• Funny -- "show work" also appears in this answer, but there the request goes in the other direction... Commented Jan 10, 2012 at 7:46
• 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. Commented Jan 10, 2012 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) Commented Jan 10, 2012 at 8:15

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.