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I am not sure how to find the above number of possible pairs. For example, for $\Sigma = \{1,2,3,4\}$, the desired number would be $6$. I see that this is half of ${4 \choose 2}$ and I am wondering if this is in fact the way to go about this: take all possible two element selections from a set of $n$ elements, and then equate choosing $(i, j)$ with choosing $(j, i)$.

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My mistake, it is in fact simply ${n \choose 2}$ –  Zvpunry May 30 '13 at 21:03
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Your comment has it right: you’re just counting pairs of distinct elements of $\{1,\dots,n\}$. –  Brian M. Scott May 30 '13 at 21:07

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${4\choose 2}$ is 6! ${n\choose 2}$ is the number of ways of choosing 2 elements from a set of n in any order. So it will just give you the number of pairs $(i, j)$ - but that is exactly what you want, as in any pair $(i,j)$ one of the elements must be greater than the other.

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