There are $\binom{n}{2}$ distances between $n$ points in $\mathbb{R}^d$. Not all of them can be chosen freely if $n$ exceeds the number $n_d = d + 1$. If $n = n_d$ we obviously have $\binom{d+1}{2}$ distances which can be chosen (more or less) independently (restricted only by the triangle inequality).

I see two ways to count $N_n^d$, the number of independent distances between $n\geq d$ points in $\mathbb{R}^d$, which is given by $nd - \binom{d+1}{2}$

The first one: $nd$ coordinates minus one translation ($d$) minus one rotation ($\binom{d}{2}$):

$N_n^d = nd - d - \binom{d}{2} = nd - (\binom{d}{2} + d) = nd - \binom{d+1}{2}$

The second one: $\binom{d+1}{2}$ distances between $d + 1$ *base points* plus $d\ (\ n - (d + 1)\ )$ distances between the remaining points and $d$ of the base points (the remaining one serving to say in which half-space with respect to the $d$ base points the point is located):

$N_n^d = \binom{d+1}{2} + d\ (\ n - (d + 1)\ ) = nd - d (d + 1) + \binom{d+1}{2} = nd - \binom{d+1}{2}$. (Is this sound?)

Observation: The binomials seem to come from two very different directions (with two seemingly different interpretations), also the term $nd$. Does this tell something deep about (Euclidean) geometry? And what?

Are there further "independent" ways to compute $N_n^d$?