# Advent calendar 2018

## 19 December

Today's number is the number of 6-dimensional sides on a 8-dimensional hypercube.

#### Show answer

#### Hide answer

Let \(f(a,b)\) be the number of \(a\)-dimensional sides on a \(b\)-dimensional hypercube. A \((b+1)\)-dimensional hypercube is formed by making
two copies of a \(b\)-dimensional hypercube and adding edges between the corresponding vertices on the two copies.
Therefore the \((b+1)\)-dimensional hypercube will contain two copies of each \(a\)-dimensional side in the \(b\)-dimensional hypercube;
it will also contain a new \(a\)-dimensional side formed by each \((a-1)\)-dimensional side of the \(b\)-dimensional hypercube.
Therefore \(f(a,b+1)=2f(a,b)+f(a-1,b)\).

(This can be seen more clearly by looking at the 1-dimensional sides (edges) and 2-dimensional sides (faces)
on a 3-dimensional hypercube (cube) and a 2-dimensional hypercube (a square).)

Using the above relation, we find that there are **112** 6-dimensional sides on a 8-dimensional hypercube.