Desargues Theorem of two triangles in perspective has symmetric order 120(why?)

Question

10 points and 10 lines construct Desargues' theorem, but since the order of the symmetric group is 120 this means we are permuting 5 elements. But I am confused to what these elements are. In total there are 6 points on both triangles in the configuration, so I am a bit lost as to how we get 120. My teacher also told us that we get 10 points from the permutation 5 choose 3(5,3)=10, which gives us the points. Anyway you can help me connect these dots?

Answer 1

You can interpret Desargues' configuration (and proove his theorem) by considering five planes in three-dimensional space (in general position). These will intersect in 10 lines (where two planes intersect) and 10 points (where three planes intersect), which can be projected onto yet another plane to give the configuration you know.

Now if you permute the labels of the planes, then you obtain a different labeling for the points and lines, but the same structure. There are $5!=120$ possible ways to label the five planes.

The $\binom53=10$ points your teacher mentioned are exactly those possible choices of three intersecting planes, since each point is the intersection of three planes. Likewise you get $\binom52=10$ lines, since each line is the intersection of two planes.

Here is a version of Desargues' configuration where the lines are labeled by the indices of two planes, and the points by the indices of three planes. You can imagine that any permutation of $\{1,2,3,4,5\}$ would lead to a combinatorially equivalent labeling