For a set x, let ${\cal S}\left$ be the set of all permutations of x. We prove in ZF several results concerning this notion, among which are the following: For all sets x such that ${\cal S}\left$ is Dedekind infinite, $\left| {{{\cal S}_{{\rm{fin}}}}\left} \right| < \left| {{\cal S}\left} \right|$ and there are no finite-to-one functions from ${\cal S}\left$ into ${{\cal S}_{{\rm{fin}}}}\left$, where ${{\cal S}_{{\rm{fin}}}}\left$ denotes the set of all permutations of x which move only finitely many elements. For all sets (...) x such that ${\cal S}\left$ is Dedekind infinite, $\left| {{\rm{seq}}\left} \right| < \left| {{\cal S}\left} \right|$ and there are no finite-to-one functions from ${\cal S}\left$ into seq, where seq denotes the set of all finite sequences of elements of x. For all infinite sets x such that there exists a permutation of x without fixed points, there are no finite-to-one functions from ${\cal S}\left$ into x. For all sets x, $|{[x]^2}| < \left| {{\cal S}\left} \right|$. (shrink)

For a set x, let S(x) be the set of all permutations of x. We prove by the method of permutation models that the following statements are consistent with ZF: (1) There is an infinite set x such that |p(x)|<|S(x)|<|seq^1-1(x)|<|seq(x)|, where p(x) is the powerset of x, seq(x) is the set of all finite sequences of elements of x, and seq^1-1(x) is the set of all finite sequences of elements of x without repetition. (2) There is a Dedekind infinite set (...) x such that |S(x)|<|[x]^3| and such that there exists a surjection from x onto S(x). (3) There is an infinite set x such that there is a finite-to-one function from S(x) into x. (shrink)

For a cardinal, we write for the cardinality of the set of permutations with finitely many non‐fixed points of a set which is of cardinality. We investigate the relationships between and for an arbitrary infinite cardinal in (without the axiom of choice). It is proved in that for all infinite cardinals, and we show that this is the best possible result.

It is proved in $\mathsf {ZF}$ (without the axiom of choice) that, for all infinite sets M, there are no surjections from $\omega \times M$ onto $\operatorname {\mathrm {\mathscr {P}}}(M)$.

It is proved in (without the axiom of choice) that for all infinite cardinals and all natural numbers, where is the cardinality of the set of permutations with exactly non‐fixed points of a set which is of cardinality.

A partition is finitary if all its members are finite. For a set A, $\mathscr {B}(A)$ denotes the set of all finitary partitions of A. It is shown consistent with $\mathsf {ZF}$ (without the axiom of choice) that there exist an infinite set A and a surjection from A onto $\mathscr {B}(A)$. On the other hand, we prove in $\mathsf {ZF}$ some theorems concerning $\mathscr {B}(A)$ for infinite sets A, among which are the following: (1) If there is a finitary (...) partition of A without singleton blocks, then there are no surjections from A onto $\mathscr {B}(A)$ and no finite-to-one functions from $\mathscr {B}(A)$ to A. (2) For all $n\in \omega $, $|A^n|<|\mathscr {B}(A)|$. (3) $|\mathscr {B}(A)|\neq |\mathrm {seq}(A)|$, where $\mathrm {seq}(A)$ is the set of all finite sequences of elements of A. (shrink)

For a set M, let |M| denote the cardinality of M. A family F is called strongly almost disjoint if there is an n∈ω such that |A∩B|<n for any two distinct elements A, B of F. It is shown in ZF (without the axiom of choice) that, for all infinite sets M and all strongly almost disjoint families F⊆P(M), |F|<|P(M)| and there are no finite-to-one functions from P(M) into F, where P(M) denotes the power set of M.

The factorial of a cardinal, denoted by, is the cardinality of the set of all permutations of a set which is of cardinality. We give a condition that makes the cardinal equality provable without the axiom of choice. In fact, we prove in that, for all cardinals, if and there is a permutation without fixed points on a set which is of cardinality, then.

The factorial of a cardinal, denoted by, is the cardinality of the set of all permutations of a set which is of cardinality. We give a condition that makes the cardinal equality provable without the axiom of choice. In fact, we prove in that, for all cardinals, if and there is a permutation without fixed points on a set which is of cardinality, then.