Number of independent distances between n points in d-dimensional Euclidean space? 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 $N_n^d$?
 A: The additional `constraints' missing 
from these  $N \choose 2$ interparticle distances  are rank constraints
on the corresponding matrix $M$ of (squared) distances. 
There is a matrix parameterization of your  space,
essentially due to Cayley. 
Consider the matrix $M$ whose ij entries arethe   squared distance
$\| p_i - p_j \|$ between  your $n$ points $p_i$
in $R^d$.  It is pretty clear that $M$ is a symmetric positive semi-definite
real matrix.   What is   less obvious is that the 
rank of $M$ is less than or equal to  $min(n-1, d)$
(  equality if and only if the points are in general position).
The converse is also
true: any symmetric n by n positive semi-definite
matrix  $M$ of rank less than or equal to $min(n-1, d)$
is realized as having its entries as the squared distances
between $n$ points in $R^d$.   
Standard formulae for the dimensions
of the space of symmetric matrices of a given rank verify
that the dimension formula is as you have   have written.
These formulae - and the method behind them --  can be found in
papers or book of Arnol'd on singularity theory. 
For a wonderful way to see this fact,
and for the  historical references to Cayley and others,
see Le problème des n corps et les distances mutuelles'',
Alain Albouy and Alain Chenciner, Inventiones mathematicae 131 pp. 151-184 (1998).
The crux of their argument is to make up an n-1 dimensional abstract vector
space ofdispostions''.  Then $M$ can be expressed as $x \circ g \circ x^t$
where $x$ is a linear map from the disposition space to $R^d$ encoding
the particle positions modulo translations, and $g$ encodes the metric on $R^d$. 
The fact that $rank(M) \le min(n-1, d)$ follows immediately.  
A: 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?
