## K-dimensional faces of an N-dimensional hypercube

How many K-dimensional faces does an N-dimensional hypercube have? The picture above shows that when N = 3, there are eight 0-dimensional faces (vertices), twelve 1-dimensional faces (edges), six 2-dimensional faces, and one 3-dimensional “face.” But what’s the answer in general?

Here’s a way to “see” it.

First, the one-dimensional case:

Now, the two-dimensional case:

The 3-dimensional case, pictured at the top of this blog, illustrates $(1 + x + 1)^3$.

Get it?

That is, the number of K-dimensional faces of the N-dimensional hypercube is the coefficient of $x^K$ in the expansion of $(1 + x+ 1)^N$, which, using the binomial theorem, is equal to ${N \choose K}2^{N-K}$.