= Solution
In a <symmetric group>, two <permutations> are in the same <conjugacy class> exactly when they have the same <cycle type>. For $S_4$ the five possibilities and their sizes are
$$
\boxed{\begin{array}{c|c|c}
\text{cycle type}&\text{representative}&\text{class size}\\ \hline
1^4&e&1\\
2\,1^2&(12)&6\\
2^2&(12)(34)&3\\
3\,1&(123)&8\\
4&(1234)&6
\end{array}}.
$$
There are $\binom42=6$ <transpositions>, three pairings into two unordered pairs, $\binom43(3-1)!=8$ three-cycles, and $(4-1)!=6$ four-cycles. Together with the identity these counts sum to $24=|S_4|$.
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