Number of non-degenerate conics in the finite projective plane

Question

I am looking to determine the number of non-degenerate conics in the finite projective plane $\mathbb{P}\mathbb{F}_q^2$ of order $q$. In Rey Casse's Projective Geometry: an Introduction, they determine that number to be $$\frac{M}{\left(\begin{array}{c}{q+1} \\ {5}\end{array}\right)}=\frac{1 / 5 !\left(q^{2}+q+1\right)\left(q^{2}+q\right) q^{2}(q-1)^{2}(q-2)(q-3)}{1 / 5 !(q+1) q(q-1)(q-2)(q-3)}= q^5-q^2.$$ $M$ denotes here the number of 5-arcs (sets of five points such that no three are colinear) in $\mathbb{P}\mathbb{F}_q^2$. I have two questions.

1) Could someone explain me where the 1/5! comes from in $M$?

2) I understand that a non-degenerate conic is determined by a 5-arc in $\mathbb{P}\mathbb{F}_q^2$. So why do we need to divide by $\left(\begin{array}{c}{q+1} \\ {5}\end{array}\right)$ to get the number of conics?

Answer 1

1. $M$ would seem to be constructed by the process:

   • Choose point $p_0$
   • Choose point $p_1\neq p_0$
   • Choose point $p_2$ not collinear with $p_0 - p_1$
   • Etc.

   So it selects ordered 5-tuples of points, and to reduce to counting unordered sets it must divide by the number of ways to order 5 points.

2. How many points does a conic pass through?