One-To-One functions

Question

Let $A$ be the set with $n$ elements and $B$ be the set with $m$ elements. How many one-to-one functions are there from $A$ to $P(B)$ (power set of $B$).
There are $n!$ total functions from $A$ to $B$.
and $(2^m)^n$ from $A$ to $P(B)$
Since $P(B)$ is a power set, there are $2^m$ elements in $P(B)$.
Therefore total one-to-one functions from $A$ to $P(B)$ are:
$P(n,2^m) = n!/(n-2^m)!$ for $n \ge 2^m$
or $P(2^{mn},2^m) = (2^{mn})!/(2^{mn} - 2^m)!$ ?
Is my reasoning correct in any case?

Answer 1

There are $2^{m}$ such elements in $\mathcal{P}(B)$. So assuming $2^{m} \geq n$ (as otherwise, no such function will be one to one), we have $P(2^{m}, n)$ such functions.

You can see this in the following way. Take $a_{1} \in A$. So a function $f$ has $2^{m}$ choices to map $a_{1}$. It then has $2^{m} - 1$ choices to map $a_{2}$. We continue this through $a_{n}$. We continue this in the following way to get:

$$\prod_{i=0}^{n-1} (2^{m} - i) = P(2^{m}, n)$$