Cardinality of the set of bijective functions on $\mathbb{N}$? I learned that the set of all one-to-one mappings of $\mathbb{N}$ onto $\mathbb{N}$ has cardinality $|\mathbb{R}|$.
What about surjective functions and bijective functions?
 A: Both have cardinality $2^{\aleph_0}$. Note that the set of the bijective functions is a subset of the surjective functions.
To see that there are $2^{\aleph_0}$ bijections, take any partition of $\Bbb N$ into two infinite sets, and just switch between them. It is not hard to show that there are $2^{\aleph_0}$ partitions like that, and so we are done.
A: In addition to Asaf's answer, one can use the following direct argument for surjective functions:

Consider any mapping $f: \Bbb N \to \Bbb N$ such that:
$$\forall n \in \Bbb N: f(2n) = n$$
Then $f$ is surjective, but for any $g: \Bbb N \to \Bbb N$ we may define $f(2n+1) = g(n)$, effectively showing that there are at least $2^{\aleph_0}$ surjective functions -- we've demonstrated one for every arbitrary function $g: \Bbb N \to \Bbb N$.
A: Choose one natural number. How many are left to choose from?
More rigorously, $$\operatorname{Aut}\mathbb{N} \cong \prod_{n \in \mathbb{N}} \mathbb{N} \setminus \{1, \ldots, n\} \cong \prod_{n \in \mathbb{N}} \mathbb{N} \cong \mathbb{N}^\mathbb{N} = \operatorname{End}\mathbb{N},$$ where $\{1, \ldots, 0\} := \varnothing$. The first isomorphism is a generalization of $\#S_n = n!$ Edit: but I haven't thought it through yet, I'll get back to you.
A: The same. It suffices to show that there are $2^\omega=\mathfrak c=|\Bbb R|$ bijections from $\Bbb N$ to $\Bbb N$. Let $P$ be the set of pairs $\{2n,2n+1\}$ for $n\in\Bbb N$. (My $\Bbb N$ includes $0$.) For each $S\subseteq P$ define 
$$f_S:\Bbb N\to\Bbb N:k\mapsto\begin{cases}
k+1,&\text{if }k\in p\text{ for some }p\in S\text{ and }k\text{ is even}\\
k-1,&\text{if }k\in p\text{ for some }p\in S\text{ and }k\text{ is odd}\\
k,&\text{if }k\notin\bigcup S\;;
\end{cases}$$
the function $f_S$ simply interchanges the members of each pair $p\in S$. Clearly $|P|=|\Bbb N|=\omega$, so $P$ has $2^\omega$ subsets $S$, each defining a distinct bijection $f_S$ from $\Bbb N$ to $\Bbb N$. Thus, there are at least $2^\omega$ such bijections. And each function of any kind from $\Bbb N$ to $\Bbb N$ is a subset of $\Bbb N\times\Bbb N$, so there are at most $2^\omega$ functions altogether. Thus, there are exactly $2^\omega$ bijections.
A: Upper bound is $N^N=R$; lower bound is $2^N=R$ as well (by consider each slot, i.e. element on $x-$axis, as having $2i, 2i+1$ two choices and each combination of such choices is bijection). So answer is $R$. 
