1*. Into how many regions do $n$ planes divide space if every three of them have exactly one common point, and no four of them have a common point?
1. The problem is solved similarly to problem 3 from § 1. To the question: into how many separate parts does $n$ lines divide the plane if every two of them have a common point, and every three do not have a common point? - The answer obtained there was: $F_{2}(n)=\frac{n(n+1)}{2}+1$.

Let now $n$ planes divide the space into $F_{3}(n)$ parts. Intersecting the space with another $(n+1)$-th plane, we will divide a row of these parts into two pieces. Since $n$ planes define $n$ lines on the $(n+1)$-th plane, the number of parts of space divided by the $(n+1)$-th plane is $F_{2}(n)=\frac{n(n+1)}{2}+1$. Therefore,

$$
\begin{gathered}
F_{3}(n+1)=F_{3}(n)+\frac{n(n+1)}{2}+1, \\
F_{3}(n)=F_{3}(n-1)+\frac{n(n-1)}{2}+1 \text { and so on. }
\end{gathered}
$$

Let us give this formula a more symmetric form. First, we transform the expression $\frac{n(n+1)}{2}+1$. It can be written as:

$$
F_{2}(n)=\frac{n(n+1)}{2}+1=\frac{n(n-1)}{2}+\frac{n}{1}+1
$$

Since $\frac{n(n-1)}{1 \cdot 2}=C_{n}^{2}, \frac{n}{1}=C_{n}^{1}, 1=C_{n}^{0}$, then for the number of regions when the plane is divided by $n$ lines, we get the formula:

$$
F_{2}(n)=C_{n}^{2}+C_{n}^{1}+C_{n}^{0}
$$

It is natural to assume that for $F_{3}(n)$ we can write a similar formula:

$$
\begin{gathered}
F_{3}(n)=C_{n}^{3}+C_{n}^{2}+C_{n}^{1}+C_{n}^{0}= \\
=\frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}+\frac{n(n-1)}{1 \cdot 2}+\frac{n}{1}+1
\end{gathered}
$$

It is not difficult to verify this for $n=0,1,2,3$. Assuming that it is valid for $n$ planes, we will prove that it remains valid for $n+1$ planes. For this, we substitute the value of $F_{3}(n)$ from formula (1) into the recursive formula (*). Then we get:

$$
\begin{aligned}
F_{3}(n+1) & =\frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}+\frac{n(n-1)}{2}+\frac{n}{1}+1+\frac{n(n-1)}{1 \cdot 2}+\frac{n}{1}+ \\
& +1=\frac{(n+1) n(n-1)}{1 \cdot 2 \cdot 3}+\frac{n(n+1)}{2}+\frac{n+1}{1}+1
\end{aligned}
$$

Thus, the formula remains valid. According to the principle of mathematical induction, formula (1) is valid for any $n$.