Determine a formula for the number of subspaces that are generated by $n$ planes that satisfy the next two conditions.
First condition: any 3 planes must intersect at a point. Second condition: any 4 planes chosen must not have a common point.
As a part of the solution, the text provides that the formula is $\frac{n^3+5n+6}{6}$ and be proven by induction (same used as when counting regions for n lines in which any three lines do not intersect at the same point)
For $n=1$ , we have the base case and two subspaces.
For $n=2$ , we have four subspaces.
For $n=3$ , this plane has to intercept the other two planes, and they have to share a point in common
For $n=4$ , we have that the planes generated by the faces of a tetrahedron satisfy the conditions. Now, I believe that all groups of 4 planes will form a tetrahedron, because it satisfies the condition. Therefore, any aditional plane must create a tetrahedron.
For $n=5$ , I am adding the plane that contains a triangle inside the tetrahedron of $n=4$. If we consider the base of the tetrahedron as a system of reference for height, each of its vertices is at a different height, otherwise a tetrahedron can not be formed. Also, none of the vertices can be located at any of the triple intersection of a plane (tetrahedron's vertices). When adding the next plane, if we were to project the edges of the shape on its base, and after we add the next plane, it has created another edge, which can also be seen as a line (even if they don't touch were they would seem to do in the projection, we know that there will be an interception)... And at that point, I believe we somehow have to use the previous exercise for lines, its formula, because each line is defining $n+1$ regions, but then we have $n$ previous lines, but I don't get to see any further than that.
Any help will be appreciated, thanks!