= Solution
For the <cardinality of total orders on a countably infinite set>, a <total order> on $\mathbb N$ is a <subset> of $\mathbb N\times\mathbb N$, so there are at most $2^{\aleph_0}=\mathfrak c$ such orders. To obtain the lower bound, associate to every <binary sequence> $(b_n)$ an order by placing the two-element blocks $\{2n,2n+1\}$ in increasing order of $n$. Inside block $n$, use $2n<2n+1$ when $b_n=0$ and reverse them when $b_n=1$.
This is a <total order> for every <sequence>. Distinct <sequences> differ on the comparison of some adjacent pair and therefore produce distinct relations. Thus <binary sequences> inject into the order <set>, proving
$$
\boxed{\mathfrak c.}
$$
The construction counts distinct orders on the labelled <set>, even though the resulting <order types> are all isomorphic to the usual natural-number order.
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