Deriving that a cube has six sides via a square and combinatorics Does there exist a derivation that a cube has 6 sides from knowing that a square has 4 edges and $4\choose2$ = 6? I was thinking maybe there exists some bijective map from any 2 given edges of a square to faces of a cube but I'm not really getting anywhere. I was thinking of this 2d --> 3d example, but I imagine if a derivation exists for this it could inductively find the number of externally touching distinct sides of a straight-edged shape (e.g. a line, square, cube, etc.) for higher dimesnions too. When looking at lower dimensions, though, a line has 2 edge facing sides but $\not\exists{x}$ s.t. $2\choose{x}$ $={4}$, so maybe you can only start the induction from the 2nd dimension? I'm pretty lost so any help would be greatly appreciated. As for my background, I've taken a class on algebra but don't have any topology knowledge, so I apologize in advance if this question is rudimentary. Thanks!
 A: You totally can.  Consider the binomial $$p + s + p$$  which is a sort of combinatorial description of a line segment.  It has a point at one end, then the segment part, then another point at the other end.  We can combine the like terms to get $$s + 2p$$ which tells us it has two endpoints, and one segment.
Now let's square that:  $$(s+2p)^2 = s^2 + 4ps + 4p^2 $$
That's a square!  The $s^2$ is the square itself, because $s^2$ is a segment squared.  The $4ps$ is the four sides because a point times a segment is just a segment.  The $4p^2$ is the four vertices because a point times a point is just a point.
(If that bugs you, just set $p=1$ so that it reduces to $s^2 +4s^1 + 4s^0$, and it still counts the number of parts: one two-dimensional square, four one-dimensional sides, and four zero-dimensional corners.)
Now let's do a cube:  $$(s+2p)^3 = s^3 + 6ps^2 + 12p^2s + 8p^3$$
There's your cube: the $s^3$ is the interior part.  The $6ps^2$ is the six faces.  The $12p^2s$ is the twelve edges.  The $8p^3$ is the eight vertices.
And yes, this works for higher-dimensional cubes also.

You can use the binomial theorem to count the number of parts of an n-cube: since
$$
\begin{align}
(s+2p)^n 
& =\sum_{k=0}^n \binom nk s^{n-k}(2p)^k \\
& = \binom n0 s^n + \binom n1 2s^{n-1}p + \binom n2 4s^{n-2}p^2 + \dots + \binom n{n-1}2^{n-1}sp^{n-1} +  \binom nn2^np^n\\
\end{align}
$$
this tells you that the number of $k$-dimensional components  is the coefficient of the $s^k$ term, which is  $$\binom n{n-k}2^{n-k}.$$
Since it's just the binomial theorem, it connects with Pascal's triangle also; Pascal's triangle is nothing but a tabular representation of the binomial coefficients $\binom nk$.  We can tabulate these cube coefficients the same way:
$$
1\\
1\quad 2 \\
1\quad 4 \quad 4 \\
1\quad 6 \quad 12 \quad 8 \\
\vdots
$$
Here the rule is that each number is the sum of the number above and to the right, and twice the number above and to the left.
A: Your question is rudimentary in some sense, but nonetheless very interesting.
I think you should start with triangles and their generalizations before moving on to squares and cubes.
Imagine a triangle in the plane (with $3$ vertices and $3$ edges). Pick a fourth point in space (not in that triangle's plane) and join it to the vertices of the original triangle. You have built a tetrahedron, with counts
$(3+1, 3+3, 1+3) = (4,6,4)$ for (vertices, edges, triangles). If you now join everything to a new point in the next (fourth) dimension you see counts $(5,10,10,5)$ for (points, edges, triangles, tetrahedra). You can continue this recursive construction. The counts form the rows of Pascal's triangle.
That is even clearer if you start the argument from dimension $0$, where the figure is a single point. Then adding a point on a line leads to $(2,1)$ for (points,edges). The next step gets you to the triangle.
There is a good argument for including a $1$ at the  beginning   to make the connection to Pascal's triangle even clearer.
To build a square from a line segment you slide it parallel to itself in the plane. That turns the counts $(2,1)$ for (points, edges) to $(4,4,1)$ for (points, edges, squares). Sliding the square into space leads to counts
$$
\text{(points, edges, squares, cubes)} = (4+4, 4+4+4, 4+2,1) = (8,12,6,1).
$$
You can continue this on through the dimensions.
