Consider a set $P$ of points and a set $B$ of subsets of $P$ called lines. A Projective plane is an ordered pair $(P,B)$ satisfying:
- There is a unique line joining any two points.
- Any two lines intersect at a unique point.
- There are at least four points, no three of which are collinear.
By showing that any two lines have the same number of points and that any two points are on the same number of lines, one concludes by a counting argument that $|P|=|B|$. If the projective plane has order $n\ge 2$, these equal $n^{2}+n+1$, and there are $(n^{2}+n+1)!$ bijections between $P$ and $B$.
- How many bijections $\phi:P\rightarrow B$ satisfy the property that each point is sent to a line containing it, i.e. for each $p\in P$ we have $p\in \phi(p)$?
- How many bijections $\psi:P\rightarrow B$ satisfy the property that each point is sent to a line not containing it, i.e. for each $p\in P$ we have $p\not\in \psi(p)$?
A bad upper bound for 1. seems to be $(n+1)^{n^{2}+n+1}$, as each point is on exactly $n+1$ lines and there are $n^{2}+n+1$ points. Therefore an equaly bad upper bound for 2. is given by subtracting that amount from $(n^{2}+n+1)!$ since every bijection satifying 2. is certainly among the many more not satisfying 1.
At the very least, how can one show that at least one bijection of each type exists? (I have checked they do exist for $n=2$ and $3$)
I have no idea yet about how to answer questions 1. and 2. However, I have had an idea to at least prove that bijections of the type described in 1. and 2. exist, and since it is too long for one (or several) comments, I will post it as an answer.
In order to demonstrate the existence of bijections of the type described in 1. and 2. the idea is to start with any bijection $\varphi_{0}:P\rightarrow B$ and then follow a finite process, at the end of which, we always obtain a bijection with the desired properties.
Using an example, I will outline the process to obtain a bijection $\phi:P\rightarrow B$ with the property that for each $p\in P$ we have $p\in \phi(p)$.
$P=\{1,2,3,4,5,6,7\}$ and $B=\left\{\begin{matrix} \{1,2,5\}& \{1,3,6\}& \{1,4,7\}& \{5,6,7\}\\ \{3,4,5\}& \{2,4,6\}& \{2,3,7\} \end{matrix}\right\}$
Notation: for simplicity, we denote a line $\{a,b,c\}$ simply as $abc$. We start with any bijection $\varphi_{0}:P\rightarrow B$, which in our example will be $$\begin{matrix} 1\stackrel{\varphi_{0}}{\mapsto} 246\\ 2\mapsto 136\\ 3\mapsto 147\\ 4\mapsto 567\\ 5\mapsto 345\\ 6\mapsto 237\\ 7\mapsto 125 \end{matrix}$$
Now, is $1\in\varphi(1)$? NO, then check: is $1\in \varphi(2)$? YES, then swap $\varphi(1)$ and $\varphi(2)$. This defines a new bijection $\varphi_{1}:P\rightarrow B$ with the property that $1\in\varphi_{1}(1)$, and $\varphi_{1}(p)=\varphi(p)$ except for $p=1,2$. That is, $$\begin{matrix} 1\stackrel{\varphi_{1}}{\mapsto} 136\\ 2\mapsto 246\\ 3\mapsto 147\\ 4\mapsto 567\\ 5\mapsto 345\\ 6\mapsto 237\\ 7\mapsto 125 \end{matrix}$$ Since $\varphi_{1}\neq\varphi_{0}$ we repeat the process with $\varphi_{1}$. That is, we know $1\in\varphi_{1}(1)$ and we check: is $2\in\varphi_{1}(2)$? YES. Then nothing to swap, and we check: is $3\in\varphi_{1}(3)$? NO, then check: is $3\in\varphi_{1}(4)$? NO, then check: is $3\in\varphi_{1}(5)$? YES, then swap $\varphi_{1}(3)$ and $\varphi_{1}(5)$. This defines a new bijection $\varphi_{2}:P\rightarrow B$ with the property that $p\in\varphi_{2}(p)$ for $p=1,2$, and $\varphi_{2}(p)=\varphi_{1}(p)$ except for $p=3,5$. That is, $$\begin{matrix} 1\stackrel{\varphi_{2}}{\mapsto} 136\\ 2\mapsto 246\\ 3\mapsto 345\\ 4\mapsto 567\\ 5\mapsto 147\\ 6\mapsto 237\\ 7\mapsto 125 \end{matrix}$$ Since $\varphi_{2}\neq\varphi_{i}$, $i=0,1$, we repeat the process with $\varphi_{2}$. This produces a bijection $\varphi_{3}:P\rightarrow B$ with the property that $p\in\varphi_{3}(p)$ for $p=1,2,3$, and $\varphi_{3}(p)=\varphi_{2}(p)$ except for $p=4,5$. That is, $$\begin{matrix} 1\stackrel{\varphi_{3}}{\mapsto} 136\\ 2\mapsto 246\\ 3\mapsto 345\\ 4\mapsto 147\\ 5\mapsto 567\\ 6\mapsto 237\\ 7\mapsto 125 \end{matrix}$$ Since $\varphi_{3}\neq\varphi_{i}$, $i\leq 2$, we repeat the process with $\varphi_{3}$. This produces a bijection $\varphi_{4}$ with the property that $p\in\varphi_{4}(p)$ for $p=2,3,4,5$, and $\varphi_{4}(p)=\varphi_{3}(p)$ except for $p=1,6$. $$\begin{matrix} 1\stackrel{\varphi_{4}}{\mapsto} 237\\ 2\mapsto 246\\ 3\mapsto 345\\ 4\mapsto 147\\ 5\mapsto 567\\ 6\mapsto 136\\ 7\mapsto 125 \end{matrix}$$ Since $\varphi_{4}\neq\varphi_{i}$, $i\leq 3$, we repeat the process with $\varphi_{4}$. This produces a bijection $\varphi_{5}:P\rightarrow B$ which will equal $\varphi_{4}$ except at the "swapping pair" $p=1,4$. That is, $$\begin{matrix} 1\stackrel{\varphi_{5}}{\mapsto} 147\\ 2\mapsto 246\\ 3\mapsto 345\\ 4\mapsto 237\\ 5\mapsto 567\\ 6\mapsto 136\\ 7\mapsto 125 \end{matrix}$$ Since $\varphi_{5}\neq\varphi_{i}$, $i\leq 4$, we repeat the process with $\varphi_{5}$. This would swap the images $\varphi_{5}(1)$ and $\varphi_{5}(4)$ of $p=1,4$ producing the bijection $\varphi_{4}$ again. What do we do in such a case?
We return to $\varphi_{4}$ and instead of swapping $\varphi_{4}(1)$ (first occurrence of a point $p$ not on his image) by $\varphi_{4}(4)$ (first image down the list which contains $p$) we go down the list to the next image containing $p$, which is $\varphi_{4}(6)$ and swap. This produces a "NEW" bijection $\varphi_{5}:P\rightarrow B$ (that is, the "new" $\varphi_{5}$ replaces the "old" one) which equals $\varphi_{4}$ except at the "swapping pair" $p=1,6$. That is, $$\begin{matrix} 1\stackrel{\varphi_{5}}{\mapsto} 136\\ 2\mapsto 246\\ 3\mapsto 345\\ 4\mapsto 147\\ 5\mapsto 567\\ 6\mapsto 237\\ 7\mapsto 125 \end{matrix}$$ But notice that this is the same as $\varphi_{3}$. Thuse, we return to $\varphi_{4}$ and instead of swapping $\varphi_{4}(1)$ and $\varphi_{4}(6)$, we go down to the following image $\varphi_{4}(p)$ containing $p$, which is our case is $\varphi_{4}(7)$. This produces a "NEW" bijection $\varphi_{5}:P\rightarrow B$ which equals $\varphi_{4}$ except at the "swapping pair" $p=1,7$. If this newly obtained $\varphi_{5}$ were to equal $\varphi_{i}$ for some $i\leq 4$ we would repeat the process. In this case the bijection $\varphi_{5}$ is $$\begin{matrix} 1\stackrel{\varphi_{5}}{\mapsto} 125\\ 2\mapsto 246\\ 3\mapsto 345\\ 4\mapsto 237\\ 5\mapsto 567\\ 6\mapsto 136\\ 7\mapsto 237 \end{matrix}$$ Since $\varphi_{5}\neq\varphi_{i}$, $i\leq 4$, we continue the checking process. But in the case of $\varphi_{5}$ there is no point $p\in P$ such $p\not\in\varphi_{5}(p)$, and so the process terminates. Thus, $\varphi_{5}$ is the desired bijection $\phi$.
All in all, quite an inefficient process, but I wanted to describe every step. However, given any bijection to start with, it seems to always terminate producing a bijection of the desired type. [I have tried several examples- e.g. beginning with a bijection $\varphi_{0}$ for which no point $p$ is in $\varphi_{0}(p)$ and the process works. HOWEVER, I may be wrong and you may find a counterexample. That would be welcomed :)]
Finally, an analog process, with the obvious "checking" modifications, seems to demonstrate the existence of bijections $\psi:P\rightarrow B$ such that for each $p\in P$ we have $p\not\in\psi(p)$.