5

If we take a point and we pass a plane through it, we see it hits it at one point once so we get $[1]$. If we take a line segment and pass a plane through where the normal line is the line from opposite corners we hit one point twice so we get $[1,1]$. Do the same with a square we see it hits a point, then 2 points simultaniously, then 1 point, so we get $[1,2,1]$. Do the same with a cube, we get $[1,3,3,1]$. Put them in a nice order, we get $$1$$$$1 \phantom{\_}1$$$$1\phantom{\_}2\phantom{\_}1$$$$1\phantom{\_}3\phantom{\_}3\phantom{\_}1$$ My question is does this pattern continue for subsequent dimensions?

Jacob Claassen
  • 868
  • 1
  • 8
  • 19

2 Answers2

3

Of course it does continue. Say, an n-dimensional cube has vertices with all coordinates either 0 or 1, and we cut it with a plane normal to one of its great diagonals. Then at step $k$ the plane passes through all points having $k$ 1's and $(n-k)$ 0's, of which there are $\binom nk$.

Ivan Neretin
  • 12,835
1

It does continue. But here's something surprising - you get binomial coefficients when you look at the planes cutting a hypercube but you do not get trinomial coefficients when you look at planes cutting a cube that can take 3 values (0,1,2) instead of just 2 (0,1). In general, the $r^{th}$ plane as you move along the main diagonal is the coefficient of $a^r$ in $(1+a+a^2+...)^n$. I made a video on this with lots of neat visualizations that covers this and many other properties. Check out - https://www.youtube.com/watch?v=KuXnrg1YpiY

Rohit Pandey
  • 6,803