Is there an Unique Generating Set of a Group when Cardinality of the set is fixed?

$A$ is a generating set of permutation-group $G$ (not a p-group), i.e. $\langle A \rangle=G$. $|A| \leq \log_2(|G|)$.

$A$ has possible maximum number of elements to generate $G$.

It means that the next possible generating set of $G$, say $C$, has cardinality $> \log(|G|)$. There is no generating set between $A$ and $C$.

Is there another set $B$ with the same cardinality of $A$ and $\langle B \rangle =G$ ?

Edit:

1. Note that every subgroup of $Sym(n)$ (symmetric group) can be generated by at most $n$ elements , this bound has been lowered, using the classification, to $[ n /2 ]$ if $n > 3$ (P. Neumann).
• For the edit what do you mean? so you don't want the maximal number of elements? And do you want non-generators with respect to this size restriction? – Paul Plummer May 27 '16 at 15:17
• So there could be a generator that seems like a non generator for small sets, but maybe there is some larger set where it is a generator. So if you only consider generating sets of a certain size it might look like some element is a non generator but when you consider your definition it isn't a non generator. – Paul Plummer May 27 '16 at 15:55
• @PaulPlummer , if i say , a minimal generating set of maximum size, does it clear things up? For example, $A, B$ have no non generator and their size is same, $C$ is another gen: set $|C|>A$ but $C$ has non generator. There is no generating set between the size of $A$ and $C$. – Jim May 27 '16 at 16:16