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 $N\_n^d$?



Number of independent distances between n points in d-dimensional Euclidean space?
Here is a little generalisation of your observation. 
Suppose we have a manfiold $M$ of dimension d with a metric. The isometry group $I(M)$ of the manifold has dimension at most $\frac{d(d+1)}{2}$. The maximal dimension of $I(M)$ is attained for $R^d$, $H^d$, $S^d$ and $RP^d$ with a constant curvature metric. Now we want to know what is the number of independent distances among $n$ points in $M^d$ with $n\ge d$. This number can be estimated from above by the reasoning identical to yours, namelly it is $nd-dim(I(M))$. 
But in reality this is only an estimation from above, because for the universal cover of $M$ the dimension will be the same as for $M$ but its group of symmetries can be larger. This is what happen for $T^n$ and $R^n$ with flat metric. 


So I wonder if the following is true. Let $M$ be a simply connected Riemannian manifold of dimension $d$. Let $Dim(M)(n)$ be the dimension of the space of independent distances among $n$ points in $M$. Is it true that $nd-Dim(M)(n)$ is the dimesnion of the group of isometries $dim(I(M))$ of $M$ for $n$ large enough?


