4D cube - I

In general, the number of $m$ D-faces of a $n$ D-cube is $2^m\binom{n}{m}.$

In this case, $n = 4$ , $m = 2$ , and the answer is $2^2\binom{4}{2} = 4\cdot 6 = \boxed{24}.$

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Proof :

Let the $n$ D-cube have vertices with coordinates $(\underbrace{\pm 1, \pm 1, \dots, \pm 1}_{n})$ .

An $m$ D-face is defined by fixing $m$ of the coordinates as either $-1$ or $+1$ . The other ones remain undefined.

We have $\binom n m$ ways to pick which coordinates these are, and $2^m$ choices for their values $+1$ or $-1$ .