Consider a compact oriented surface of genus $g$ with $k$ boundary components, ordered from $1$ to $k$. Next remove $n_i\geq 0$ points from the $i$-th boundary component, for $i=1,...,k.$ For which sequences $(g,n_1,...,n_k)$, the above surface admits a hyperbolic metric whose $i$-th component is bounded by $n_i$ infinite geodesic arcs (going to $n_i$ points at infinity)? $n_i=0$ means that the $i$-th component is a geodesic loop (or, alternatively, a puncture).
The answer is yes for all $n_i\geq 3$ but I am not sure about other cases.
There is indeed a pretty simple way to derive necessary and sufficient conditions on $(g,n_1,...,n_k)$ for existence of such a hyperbolic structure on your surface $S$.
To derive a necessary condition, let $S$ be your surface equipped with such a hyperbolic structure. Let $DS$ be obtained by doubling $S$ across the boundary, in other words $DS$ is the quotient of a pair of copies of $S$ by identifying the two copies of the boundary using the identity map. The surface $DS$ is a surface of some finite genus $g'$ and some finite number of punctures $p'$, and one can work out formulas for $g'$ and $p'$ in terms of $(g,n_1,...,n_k)$ (I'll indicate more details below of how one does this). Furthermore, by doubling the hyperbolic structure on $S$ one sees that $DS$ possesses a hyperbolic structure, i.e. a complete hyperbolic metric.
Just as for compact surfaces, existence of a complete hyperbolic metric on a punctured surface such as $DS$ is equivalent to saying that $DS$ has negative Euler characteristic: $$\chi(S) = 2 - 2g' - p' < 0 $$ It turns out that this necessary condition is also sufficient, because for any complete hyperbolic structure on $DS$, under the embedding $S \hookrightarrow DS$ one can straighten the boundary of $S$ to get the desired hyperbolic structure on $S$.
The number of punctures on $DS$ is simply $$p' = \sum_{i=1}^k n_i $$ because there is exactly one puncture on $DS$ for each boundary point that was removed to form $S$.
To compute genus, its easier to work with Euler characteristic, and we have $$\chi(S) = 2 - 2g - k - \sum_{i=1}^k n_i $$ The disjoint union of two copies of $S$ has twice the Euler characteristic: $$\chi(S \cup S') = 4 - 4g - 2k - 2 \sum_{i=1}^k n_i $$ Identifying two circle components of the boundaries does not change Euler characteristic, but identifying two arc components reduces the Euler characteristic by $1$: $$\chi(DS) = 4 - 4g - 2k - 3 \sum_{i=1}^k n_i + \#\{i \mid n_i=0\} $$ Since $\chi(DS) = 2 - 2g' - p'$, and since we have a formula for $p'$, we can solve for $g'$.