Problem: Let $S = \{1, \dots, n\}$.
Given a bijection $f \colon S \to S$ an \emph{orbit} of $f$ is a
set of the form $\{x, f(x), f(f(x)), \dots \}$ for some $x \in S$.
We denote by $c(f)$ the number of distinct orbits of $f$.
For example, if $n=3$ and $f(1)=2$, $f(2)=1$, $f(3)=3$,
the two orbits are $\{1,2\}$ and $\{3\}$, hence $c(f)=2$.

Given $k$ bijections $f_1$, \dots, $f_k$ from $S$ to itself, prove that
\[ c(f_1) + \dots + c(f_k) \le n(k-1) + c(f) \]
where $f \colon S \to S$ is the composed function $f_1 \circ \dots \circ f_k$.

Solution: Most motivated solution is to consider $n - c(f)$ and show this is the transposition distance.
Dumb graph theory works as well.