If we arrange $n$ hyperplanes in a $d$-dimensional space such that the number of regions is maximized, we obtain
$$R_d(n) = \sum\limits_{i = 0}^d\binom{n}{i}$$
This is a standard result that you can find in nearly any introductory combinatorics book. This is an outline of the proof:
The following recurrence relation describes this problem:
$$R_d(n) = R_d(n-1) + R_{d-1}(n-1)$$
To see why this is true, imagine taking $n-1$ hyperplanes in general position, then adding the $n^{th}$ hyperplane to the group. If we consider only this hyperplane, each intersection with the other hyperplanes is a "hyperline" in this $(d-1)$-dimensional hyperplane, which will make $R_{d-1}(n-1)$ regions on that plane.
After you convince yourself that each of these regions created on the $n^{th}$ hyperplane adds one to the original $d$-dimensional space, the recurrence is clear. This leaves us with the following:
$$\sum\limits_{i = 0}^d\binom{n}{i} = \sum\limits_{i = 0}^d\binom{n-1}{i} + \sum\limits_{i = 0}^{d-1}\binom{n-1}{i}$$
$$= \sum\limits_{i = 0}^d\binom{n-1}{i} + \sum\limits_{i = 1}^{d}\binom{n-1}{i-1}$$
$$= \binom{n-1}{0} + \sum\limits_{i = 1}^d\binom{n-1}{i} + \binom{n-1}{i-1}$$
Apply Pascal's formula, and note that $\binom{n-1}{0} = \binom{n}{0}$
$$= \binom{n}{0} + \sum\limits_{i = 1}^d\binom{n}{i}$$
$$= \sum\limits_{i = 0}^d\binom{n}{i}$$
Because the recurrence is satisfied and the initial values match, the given summation will work (this is a very lazy way to show it holds).
For circles in the plane, the number of regions is given by A014206.
$$R(n) = n^2-n+2$$
This is easy to show by induction. The base case $n=1$, should be clear. We will assume that
$$R(n) = n^2-n+2$$
If we add another circle, then it intersects each other circle twice, creating $2$ regions (overlap, and the non-overlap area inside the new circle). As we currently have $n$ circles, adding the next will create $2n$ more regions:
$$R(n) + 2n = n^2 + n + 2 = (n+1)^2 - (n+1) + 2 = R(n+1)$$
So we are done by induction.
I don't think the statement about the tetrahedra is true. If you have just two hyperplanes, it does not hold, nor even with three. Perhaps there is further detail in your textbook?
Let the lines end at the rim of the pizza. Not counting the exterior of the pizza we have, according to Euler's formula,
$$f-e+v=1\ ,$$
where $f$ is the number of faces, $e$ the number of edges, and $v=v_3+v_4$ the number of vertices in the resulting configuration. If no three cuts go through the same point there are $v_3=2n$ vertices of degree $3$ along the rim of the pizza and $v_4$ vertices of degree $4$ in its interior. From $2e=3v_3+4v_4$ it follows that $e=3n+2v_4$. Inserting this into $(1)$ we therefore obtain
$$f=3n+2v_4-(2n+v_4)+1=v_4+n+1\leq{n\choose 2}+{n\choose 1}+{n\choose0}\ ,$$
with equality iff we take care that all ${n\choose2}$ intersection points of the cuts are within the pizza.
Best Answer
With four lines, there are seven different ways, i.e. numbers of regions into which a bounded space can be divided by four lines. The minimum number of regions is $4+1=5$, produced when no lines intersect within the space.
But every number of regions from $5$ to maximum $11$ is possible. For beginning with $4$ non-intersecting lines within the space, we can add $1, 2, 3$ regions by redrawing one line so as to cut first one, then two, then three other lines, as in the top row of the figure.
Then we add a $4th$ and $5th$ region by redrawing the third line so as to cut one and then both of the first two lines, as in the second row of the figure.
Finally, we get a $6th$ region by making the second line cut the first, as in the third row.
Thus to the original five regions we have added$$3+2+1=6$$Generalizing, since the minimum number of regions is $n+1$, and the maximum is$$\frac{n^2+n+2}{2}$$then the number of ways will be$$\frac{n^2+n+2}{2}-n=\frac{n^2-n+2}{2}$$
And since$$\frac{n^2-n+2}{2}=\frac{(n-1)n}{2}+1$$we see that the number of ways $n$ lines can divide the space is equal to the $(n-1)th$ triangle number plus $1$.
Note: This seems to be something more general than a geometric problem. If the space is bounded, no pair of non-intersecting lines need to be parallel. Nor do the line segments need to be straight, provided that no two intersect more than once within the space. And perhaps the only condition on the bounded space is that it be concave from within?
Correction: If the plane is unbounded, and "how many ways" means "how many arrangements," as arrangement is explained in the comments on OEIS A241600 referenced by @Daniel Mathias, then the above is not a suitable answer to the question posted, and there do appear to be nine arrangements of four lines.
The first two rows have parallels, the third does not. There is one 3-line concurrence in the second row, and a 3-line and 4-line concurrence in the third. The number of regions, left to right and top to bottom, is$$5, 8, 9, 9, 10, 10, 8, 10, 11$$Unlike the situation as I first understood it, there are gaps and repetitions in the number of regions produced by the different arrangements. This seems to make the determination of $P$ as a function of $n$ a more difficult task.
Correction continued: $P$ for $n=5$
The figures shows twenty-one arrangements for two or more lines parallel when $n=5$. In each row (except the last, which is actually two rows of two each) four lines keep a given position throughout the row as a fifth line shifts its position through the essentially different arrangements possible. After the first row showing one arrangement each for five and four parallels, the second row has only three parallels, the third has two pairs of two, and the fourth and fifth only two.
Next we can see by a single figure below the arrangements possible when no lines are parallel. Again we suppose the four lines $AB$, $AC$, $FB$, $FD$ given in position. $G$ is any point not collinear with any two of the six intersection points $A$, $B$, $C$, $D$, $E$, $F$. A fifth line through $G$ can pass through and among the six points in $6+7=13$ different ways.
But if $G$ is collinear with a line $BE$, $DC$, or $AF$, it can be seen even without another figure that the number of possible arrangements will be only $5+6=11$.
And finally, we have one arrangement each for four and five concurrent lines.
Adding them up,$$1+1+5+4+10+13+11+1+1=47$$in agreement with OEIS A241600.