Yes there is a relationship, the Euler characteristic:
For a 2-dimensional orientable manifold with boundaries embedded in $\mathbb{R}^3$, the Euler characteristic is
$\chi = V - E + F = 2 - 2g - b$
where $V$ is the number of vertices, $E$ is the number of edges, $F$ is the number of faces, $g$ is the genus of the manifold, and $b$ is the number of borders of the manifold.
For example, a disk has no holes, and one border, so its Euler characteristic is $2 - 2(0) - 1 = 1$, thus we know that $V - E + F = 1$.
On a torus, $g = 1$ and $b = 0$, thus $\chi = 2 - 2(1) - 0 = 0$, so $V - E + F = 0$
The point being that this number is independent of the exact triangulation, it only depends on the topology of the manifold.