Given $4$ equidistant points, and the question "how many line segments connect them?", you could rephrase the question to "how many pairs are there in a tuple of 4?" ("pairs" because a line segment is defined by a pair of points).
The answer would be $\frac{4!}{2!}(4-2)!=6$. But if you ask a child to connect the 4 points, they will draw a square. The two lines crossing the interior of the square are not considered. So the combinatoric formula given above does not actually compute the number of edges in a square.
With a cube we have one base, $8$ vertices, and $12$ edges. With a tesseract, there are $16$ vertices and $32$ edges. Is there a formula to compute the number edges of these shapes?
When the dimension of the object is $n$ then every vertex has $n$ neighbours. Since there are $2^n$ vertices, we get $$\frac{1}{2}n2^n=n2^{n-1}$$. The one half is there not to count edges twice.