= Solution
<Burnside lemma> says that the number of orbits of a finite group $G$ on a finite set $X$ is
$$
|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|.
$$
Indeed, double-count the set $\{(g,x):gx=x\}$. Counting first by $g$ gives the numerator. Counting first by $x$ gives $\sum_x|G_x|$; each orbit $O$ contributes $|O||G_x|=|G|$ by the <orbit-stabilizer theorem>, proving the formula.
Let $\pi(g)=|X^g|$ be the character of the <permutation representation>. Then
$$
\langle\pi,\pi\rangle
=\frac1{|G|}\sum_g|X^g|^2
$$
is, by Burnside's lemma, the number of orbits on $X\times X$. A <two-transitive group action> has exactly two such orbits: the diagonal and the ordered pairs of distinct points. Hence $\langle\pi,\pi\rangle=2$. Transitivity also gives $\langle\pi,1_G\rangle=1$. If $\pi=1_G+\sum m_i\chi_i$ is its <irreducible character>[irreducible character decomposition], then $2=1+\sum m_i^2$, so exactly one nontrivial irreducible character occurs, with multiplicity one. Thus the permutation character has precisely two distinct irreducible summands.
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