Into how many parts at most is a plane cut by $n$ lines? Into how many parts is space divided by $n$ planes in general position?
My approach:
$$p(n+1)=p(n)+n+1$$ $$s(n+1)=s(n)+p(n)$$
This solution is given by Engel (pg. 40). It starts with anthe observation that the regions into which the plane is divided by $n$ lines are defined by their vertices which are the points of intersection of the given lines.
We may assume that none of the lines is horizontal; if one is, rotate the plane by a small angle. This ensures that the regions are of two sorts: some regions (finite or not) have the lowest vertex, others do not. Every point of intersection serves as the lowest vertex of exactly one region. Therefore, for $n$ lines, there are $\binom{n}{2}$ regions of the first sort. There are $n+1$ regions of the second sort, which becomes apparent when a horizontal line is drawn that crosses all n lines below all of their "legitimate" intersection points.
Therefore, $$L_n=\binom{n}{2}+\binom{n}{1}+\binom{n}{0}$$
The latter expression can be easily generalized to a 3D problem wherein the question is about the number of regions into which n planes divide the space. The answer is
$$\binom{n}{3}+\binom{n}{2}+\binom{n}{1}+\binom{n}{0}$$
How? Also how to prove that if S(n) is the space part , then at least $\frac{2n-3}{4}$ tetradehra will exist?
What will happen if instead of a line, a circle cuts the plane?