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?