My analysis leads to different answers from the above, and to some references. **Edit:** I get different answers because I changed the question without realizing it. I'm working from the collection of pointed metric balls, i.e., metric balls with a distinguished center. Gabe asked the question about the collection of unpointed metric balls, noting that the same set can be a metric ball with two different centers. That seems more complicated, although I would suggest working from the pointed solution.

Let's let $\{1,2,\ldots,n\}$ be the points in $X$.
The distances from $i$ to the other points in the set $X$ induce a strict weak ordering of those points by their distance from $i$. This has the same information as the set of metric balls with center $i$. Thus the information in the metric is given by all of the comparisons between the distances $x_{(i,j)} = d(i,j)$ and $x_{(i,k)} = d(i,k)$. You can express these relations by a hyperplane arrangement in $\mathbb{R}^{n(n-1)/2}$, where the hyperplanes are given by the equations $x_{(i,j)} = x_{(i,k)}$.

You might also think about the triangle inequalities satisfied by all of the distances, and the fact that the distances are all positive numbers. The set of feasible distance vectors is called the "metric cone". Although the metric has an interesting combinatorial structure, it looks like it matters for nothing in this particular question: The hyperplanes all meet at an interior point of the cone in which the distances are all equal. In other words, you can always add a constant distance $h \gg 0$ to all of the distances without changing any of the metric balls, so that the triangle inequality and positivity of distance become irrelevant.

If a hyperplane arrangement has the property that all hyperplanes are given by setting two coordinates equal, then the arrangement is called "graphical". The coordinates correspond to the vertices of a graph $G$, and the hyperplanes correspond to the edges. The hyperplane arrangement gives you a partially ordered set of chambes and other faces, ordered by inclusion. This poset has a lot of properties and there are techniques to compute the number of faces from $G$. In our case, $G$ is the line graph $T_n$ of the complete graph $K_n$, which is sometimes confusingly called a triangular graph.

For $n=3$, my answer is that there are 13 different types of metrics, not 7, corresponding to the hyperplane arrangement $x=y$, $y=z$, $x=z$ in $\mathbb{R}^3$. In other words, I count 13 types of triangles: 6 scalene, 3 short-base isosceles, 3 long-base isosceles, and 1 equilateral.

Some of the types of metrics lie in chambers, meaning generic metrics in which $d(i,j) \ne d(i,k)$ for all $i$, $j$, and $k$. It is a theorem that the number of chambers of the graphical arrangment $A(G)$ of a graph $G$ is $|\chi_G(-1)|$, where $\chi_G$ is the chromatic polynomial of $G$. (See this excellent review by Richard Stanley.) So I asked Maple to compute

```
ChromaticPolynomial(LineGraph(CompleteGraph(n)),q)
```

for small values of $n$, and I got the following answers:

$$\chi_{T_3}(q) = q(q-1)(q-2)$$
$$\chi_{T_4}(q) = q(q-1)(q-2)(q^3-9^2+29q-32)$$
$$\chi_{T_5}(q) = q(q-1)(q-2)(q-3)(q-4)(q^5-20q^4+170q^3-765q^2+1804q-1764).$$

Evaluating at $q=-1$, I get that there are 1, 6, 426, and 542880 generic types of metrics on 2, 3, 4, and 5 points. This sequence is not in the Encyclopedia of Integer Sequences, although possibly it should be. I think that you can obtain the total number of faces of a graphical arrangement $A(G)$ from the Tutte polynomial of $G$, and therefore the total number of types of metrics, but I did not do the calculation.