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 $\operatorname {\mathrm {\mathscr {P}}}(M)$.

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.

We write $\mathcal {S}_n(A)$ for the set of permutations of a set A with n non-fixed points and $\mathrm {{seq}}^{1-1}_n(A)$ for the set of one-to-one sequences of elements of A with length n where n is a natural number greater than $1$. With the Axiom of Choice, $|\mathcal {S}_n(A)|$ and $|\mathrm {{seq}}^{1-1}_n(A)|$ are equal for all infinite sets A. Among our results, we show, in ZF, that $|\mathcal {S}_n(A)|\leq |\mathrm {{seq}}^{1-1}_n(A)|$ for any infinite set A if ${\mathrm {AC}}_{\leq n}$ is (...) assumed and this assumption cannot be removed. In the other direction, we show that $|\mathrm {{seq}}^{1-1}_n(A)|\leq |\mathcal {S}_{n+1}(A)|$ for any infinite set A and the subscript $n+1$ cannot be reduced to n. Moreover, we also show that “ $|\mathcal {S}_n(A)|\leq |\mathcal {S}_{n+1}(A)|$ for any infinite set A” is not provable in ZF. (shrink)