= Solution
A <composition series> is a chain $1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_r=G$ with simple nontrivial factors. Every <finite group> has one, by recursively choosing a maximal proper <normal subgroup>. The <Jordan–Hölder theorem> states that any two such series have the same length and the same multiset of simple factors up to isomorphism.
For a proof induct on $|G|$. Let $M,N$ be the penultimate terms in two series. If $M=N$, apply induction there. Otherwise their maximal normality gives $MN=G$. Set $D=M\cap N$. The second isomorphism theorem gives
$$
M/D\cong G/N,\qquad N/D\cong G/M,
$$
so these quotients are simple. Choose a <composition series> of $D$ and extend it once to $M$ and once to $N$. Induction in the smaller <groups> $M,N$ compares these with the respective original prefixes. Appending $G/M$ or $G/N$ gives in both cases the factors of $D$ together with $G/M,G/N$. This completes the proof, including the case $D=1$.
\b[Exactly one:] take $S_5$. Its only nontrivial proper <normal subgroup> is $A_5$. Indeed a normal $N$ has $N\cap A_5$ either $1$ or $A_5$ by the supplied simplicity. In the first case $N$ injects into $S_5/A_5\cong C_2$, so a nontrivial $N$ would be a <normal subgroup> of order two and hence central. But $S_5$ has trivial center: conjugating every transposition trivially forces preservation of each unordered pair, and therefore of every point. Thus the unique series is $1\triangleleft A_5\triangleleft S_5$.
\b[Exactly two:] take $A_5\times C_2$. A <normal subgroup> containing $A_5\times1$ is that subgroup or the whole product. For a <normal subgroup> $N$ meeting $A_5\times1$ trivially, commutators of its elements with $A_5\times1$ vanish; its first coordinates therefore lie in $Z(A_5)=1$. Hence $N\subseteq1\times C_2$. The two nontrivial proper <normal subgroups> are exactly the two factors, each simple, so the two series pass respectively through $A_5\times1$ and $1\times C_2$.
\b[More than two:] take $C_{120}$. A <cyclic group> has exactly one subgroup of each divisor order, and composition steps have prime indices. The orders of those indices are the distinct permutations of $2,2,2,3,5$, giving
$$
\boxed{5!/3!=20\text{ composition series}.}
$$
All three example <groups> have order $120$.
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