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I am trying to solve the following problem from a textbook:

Let $S$ be a set with cardinality $2m$. Give a closed-form expression for the number of equivalence relations on $S$ where every equivalence class has two elements.

My approach thus far: Every equivalence relation partitions $S$ into $m$ different subsets/classes. Thus, counting the number of equivalence relations is equivalent to counting the number of ways of choosing $m$ different objects in a $2m$-element set. And this is given by $\binom{2m}{m}$.

Is my approach correct? If not, how can I solve this problem? I would appreciate any help on this!

John
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1 Answers1

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Unfortunately your approach chooses $m$ elements from a set of $2m$ elements, but it doesn't specify who their pairs would be.

To solve this: Choose the first class of two elements. the order of choice doesn't matter so divide by two. Choose the second class of two elements from what is left. Again divide by two. Continue this way until there are no more elements. Then you have to divide by the order of the pairs, which also didn't matter.

This way you get $[2m\cdot(2m-1)/2]\cdot[(2m-2)\cdot(2m-3)/2]...[2\cdot1/2]/m!$ so this is given exactly by $$\frac{(2m)!}{2^mm!}= {2m\choose{2,2,2,...,2}}/m!$$

NL1992
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