In this paper, we consider certain cardinals in ZF (set theory without AC, the axiom of choice). In ZFC (set theory with AC), given any cardinals $\mathscr{C}$ and $\mathscr{D}$, either $\mathscr{C} \leq \mathscr{D}$ or $\mathscr{D} \leq \mathscr{C}$. However, in ZF this is no longer so. For a given infinite set $A$ consider $\operatorname{seq}^{1 - 1}(A)$, the set of all sequences of $A$ without repetition. We compare $|\operatorname{seq}^{1 - 1}(A)|$, the cardinality of this set, to $|\mathscr{P}(\mathscr{A})|$, the cardinality of the power set of $A$. What is provable about these two cardinals in ZF? The main result of this paper is that $ZF \vdash \forall A(|\operatorname{seq}^{1 - 1}(A)| \neq|\mathscr{P}(\mathscr{A})|)$, and we show that this is the best possible result. Furthermore, it is provable in ZF that if $B$ is an infinite set, then $|\operatorname{fin}(B)| < |\mathscr{P}(B)|$ even though the existence for some infinite set $B^\ast$ of a function $f$ from $\operatorname{fin}(B^\ast)$ onto $\mathscr{P}(B^\ast)$ is consistent with ZF.