= Solution
The <symmetric group> $S_5$ has <cardinality> $5!=120=3\cdot40$, so its <Sylow subgroups> for $p=3$ have <cardinality> $3$. One example is
$$
\boxed{\langle(123)\rangle=\{e,(123),(132)\}.}
$$
Every nonidentity <permutation> of <order of a group element> $3$ in $S_5$ is a single <permutation cycle> of length $3$: two disjoint length-three <permutation cycles> would require six letters. Choose the three moved letters in $\binom53$ ways, then one of their two length-three <permutation cycles>, giving $20$ such elements. Each subgroup of <cardinality> $3$ has exactly two nonidentity elements, and distinct such subgroups have no nonidentity element in common. Consequently
$$
\boxed{n_3(S_5)=\frac{\binom53\,2}{2}=10.}
$$
This also satisfies the numerical <Sylow theorem> restrictions $10\mid40$ and $10\equiv1\pmod3$.
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