I was looking through images of hypercubes in higher dimensions when I came upon this picture:
https://i.stack.imgur.com/o9mmC.jpg
I couldn’t help but notice that the squares facing the viewer seem to form Pascal’s triangle. In 2D, there is just 1 square. In 3D, there is 1 and 1 (1 1) facing the viewer. In 4D, there is 1, then 2 overlapping, then another 1 (1 2 1). The same with 5D (1 3 3 1) but more difficult to see because of the lines.
Is this just a coincidence?
No, it is not a coincidence. Let me explain.
An $n$-hypercube is formed by taking all of the points in $n$-dimensional space with coordinates of the form $(\pm1,\pm1,\ldots,\pm1)$, and connecting those that differ by a single entry. To show it in a 2D screen, we might choose to first divide it into “front-facing” squares having all but the first two coordinates equal, and then, layer these squares depending on the number of plus signs on the rest of the coordinates. This has the advantage that any two adjacent vertices are either on the same front-facing square, or on adjacent layers. This is also the arrangement your image shows.
As to why this generates Pascal’s Triangle: The $k$-th layer of the image of an $n$-hypercube layer will have as many squares as ways to choose $k$ plus signs from $n-2$. This may be calculated by the Binomial Coefficient $\binom{n-2}{k}$, which you might also recognize as the $k$-th entry of the $(n-2)$-th row of Pascal’s Triangle (if we start indexing from $0$ up). This is where these numbers are coming from.