For the sake of simplicity of exposition, let $G$ be a non-cyclic finite group which can be generated by two elements. Let us first consider whether an element $x \in G$ can be a member of two-element generating set. In the contrary case, we have 
$\langle x, y \rangle < G$ for any $y \in G$. Continuing the contrary case, if we set $\mathcal{M}(x)$ to be the union of the maximal subgroups containing $x$, we see that 
$\langle x,y \rangle \leq \mathcal{M}(x)$ for each $y \in G$, so that in particular 
$G = \mathcal{M}(x).$ 

On the other hand, if $\mathcal{M}(x) = G$, then we do have $\langle x, y \rangle < G$ for each $y in G$. Hence $x \in G$ is a member of a two-element generating set for $G$ if and only if $\mathcal{M}(x) < G.$ To make more explicit, if 
$y \in G \backslash \mathcal{M}(x)$, we see that $\langle x,y \rangle$ is not contained in any maximal subgroup of $G$ ( for that maximal subgroup would contain $x$, so would be contained in $\mathcal{M}(x)$, which it isn't as even $y$ isn't). 

So it is actually (in principle) possible to give a precise count of the number of (different in the sense of the question) two element generating sets of $G$.

First, determine those $x$ for which $\mathcal{M}(x) < G$ (assuming $G$ is two-generated, there will be some such $x$). Let $\mathcal{G}$ be the set of such $x$.
Then the number of different generating pairs for $G$ is 
$$\frac{\sum_{x \in \mathcal{G}}|G \backslash \mathcal{M}(x)|}{2},$$ since we obtain each generating pair $\{x,y\}$ twice, once from $y \in G \backslash \mathcal{M}(x)$ and once from $x \in G \backslash \mathcal{M}(y).$

The idea to generalize this to groups which are $k$-generated, but not $k-1$-generated should be clear. Given $k-1$ elements $x_{1},x_{2}, \ldots, x_{k-1}$, it can be extended to a $k$-element generating set if and only if $\mathcal{M}(\langle x_{1}, x_{2}, \ldots , x_{k-1} \rangle) < G,$ where $\mathcal{M}(\langle x_{1}, x_{2}, \ldots , x_{k-1} \rangle)$ is the union of those maximal subgroups of $G$ which contain $\langle x_{1},x_{2}, \ldots, x_{k-1} \rangle.$