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Consider the proportion

$\displaystyle\frac{a}b=\frac{c}d=\frac{e}f=\frac{g}h$.

It is equivalent to

$\displaystyle\frac{f}e=\frac{b}a=\frac{h}g=\frac{d}c$

in the sense that the fractions of one are those of the other, or all are inverted. Under that relation, how many equivalence classes are there (of proportions including $a,b,c,d,e,f,g,h$)? Well, you can construct them and count as you go, or you can consider all possible proportions and mod out by the equivalence:

Method 1: constructing inequivalent proportions:

The first element (first fraction, numerator) can arbitrarily be chosen as $a$ (i.e., this helps to pick a representative, but not a class). There are then $7$ choices for the numerator. Anything can then be chosen as the next number, but there are $2$ choices of where to put it: in a numerator or a denominator. Then there are $5$ choices of the counterpart to that third number. And so on: the number of non-arbitrary choices made in constructing an equivalence class is $7\times2\times5\times2\times3\times2\times1$.

Method 2: modding out:

There are $8!$ proportions. There are $4!$ ways to order the fractions and $2!$ choices as far which numbers are numerators, so there are $\frac{8!}{4!2!}$ classes.


I'm not seeing a direct connection between these two formulas (expressions). I would have thought, since they represent the same model (and, really, the same way of looking at the model, just worded differently), that the two formulas given my these two methods would be easy to relate to one another. I mean, sure, you can do arithmetic to get from one to the other:

$\frac{8!}{4!2!}=\frac{8!}{4\times3!2!}=\frac{7!}{3!}=7\times2\times5\times2\times3\times2\times1$,

but that requires splitting up the $4!$ into $4\times3!$ (the $4$ goes with the $2!$ to get divided into the $8$, whereas the $3!$ gets divided into the other numbers). That seems so... ugly. Am I wrong for expecting there to be some nice way of seeing the connection between these two formulas? Or is there one I'm not seeing?

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    Similar in nature to http://math.stackexchange.com/questions/2237, but not a duplicate.2011-09-09
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    They are not exactly equivalent: the first is true if $a=c=e=g=0$ and $bdfg\neq 0$, while the second is not sensible in that situation; and conversely, the second is true if $b=d=f=g=0$ and $aceg\neq 0$, while the first is not sensible in that case.2011-09-09
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    I don't understand what you are trying to count; but your method 1 does not seem to take into account the possibility that you pick the same fractions but in different order.2011-09-09
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    @Arturo: It does: that’s why he arbitrarily chooses $a$ as the first element and says that there are $7$ ways to choose its mate. Then he picks (say) the first available letter in alphabetical order to belong to another pair; he now has $5$ ways choices for its mate and $2$ for orientation of the pair. Again he picks (say) the first free letter in alphabetical order to belong to another pair, chooses one of the $3$ remaining letters for its mate, and chooses one of the $2$ orientations. This leaves one pair, whose orientation must still be specified.2011-09-09
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    @Brian: I'll take your word for it, since I still don't quite understand just what it is that is being counted... )-:2011-09-09

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