Two equivalent formulations of the axiom of choice are:

- Every family $(X_i)_{i \in I}$ of pairwise disjoint nonempty subsets of a set $X$ has a choice function.
- Every family $(X_i)_{i \in I}$ of nonempty subsets of a set $X$ has a choice function.

However, the usual proof of the implication (1) → (2) replaces the set $X$ with the set $X \times I$ and extracts a choice function for $(X_i)_{i \in I}$ from a choice function for the pairwise disjoint family $(X_i\times\lbrace i \rbrace)_{i \in I}$ of nonempty subsets of $X \times I$. Since $X \times I$ can be much larger than $X$, there is no reason to believe that (1) → (2) for a fixed set $X$.

For a given set $X$, (2) has a maximal instance where the family $(X_i)_{i \in I}$ consists of all nonempty subsets of $X$. We therefore see that (2) is equivalent to:

- There is a choice function $\mathcal{P}(X)\setminus\lbrace\varnothing\rbrace\to X$.
- The set $X$ is wellorderable.

For a given set $X$, (1) is equivalent to:

- Every surjection $q:X \to Y$ has a right inverse.
- Every equivalence relation on $X$ has a transversal.

It appears that (2) is indeed stronger than (1) for a fixed set $X$ and I feel that this should be well known, but I don't recall a model of ZF (or ZFA) where some set $X$ satisfies (1) but not (2). Does anyone know such a model? A model of ZF where $X = \mathbb{R}$ satisfies (1) but not (2) would be most interesting.