Permutations and Cross-ratios Pick four distinct numbers, list all 24 permutations, and compute the cross-ratio of each permutation. Show that at most six numbers have occurred, given by the cross-ratio group
$y, \frac{1}{y}, 1-y, 1-\frac{1}{y}, \frac{1}{1-y}, \frac{y}{y-1}$.
 A: In the first sentence, the instructor makes the unreasonable request (which may be a command, in the way that classes are often organized) to perform a long, repetitive and largely useless computation.  This is not something that people are likely to spend time doing in an online math site, and I am sorry to hear that this kind of time waste is being inflicted on students. 
For the command in the second sentence, the long computation will have produced 24 numbers, and many of those numbers will be equal to each other.  If you list the numbers and remove duplicates (which is the same as to read the numbers "as a set" instead of a list), there is a theorem about cross-ratios saying that the number of possible values is at most six, and is exactly six except for a few exceptional starting values of $y$.  So the problem appears to be a tedious computation to illustrate a theorem. 
As posted, this looks like a terrible choice of problem by the instructor, but if you really meant

Show that if you pick four distinct numbers, list all 24 permutations, and compute the cross-ratio of each permutation, then at most six numbers have occurred, given by the cross-ratio group
  $y, \frac{1}{y}, 1-y, 1-\frac{1}{y}, \frac{1}{1-y}, \frac{y}{y-1}$.

that is a more standard and instructive exercise.
Some actually useful and important facts behind the question:


*

*substituting functions of the form $\frac{ax+b}{cx+d}$ into each other produces a function of the same kind (these are called "linear fractional" and are a special case of rational functions, which are ratios of polynomials). 

*starting from $f(x)=1-x$ and $g(x)=1/x$, there are exactly six different linear fractional functions that can be formed by a chain of substitutions, and they are the six functions (of $y$, using the question's notation) listed in the second sentence.

*This six is the same six as the number of permutations of three objects, more exactly any (fixed) three of the points.


The reason for "at most six" in the problem is that the values of the six different rational functions can coincide for specific arguments, such as $y = (1-y)$ for $y = \frac{1}{2}$.
A: HINT: The stabilizer of $\frac{(a-c)(b-d)}{(b-c)(a-d)}$ contains at least $(a\,c)(b\,d)$ and $(a\,b)(c\,d)$.
