Solution
Ten Planes
Show the problem again
Ten planes in space are in general position: every two meet in a line, every three meet in a single point, no four share a point, and no two of the intersection lines are parallel. Into how many regions do they divide space? Give an integer.
Worked solution
The answer is 176. When the $k$-th plane is added, the previous $k - 1$ planes cut it along $k - 1$ lines in general position, which divide that plane into $1 + (k-1) + \binom{k-1}{2}$ regions. Each of these planar regions splits a spatial region in two. So $R(n) = R(n-1) + 1 + (n - 1) + \binom{n-1}{2}$, giving $R(n) = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \binom{n}{3}$. For $n = 10$: $1 + 10 + 45 + 120 = 176$.
Source: Classic exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.