= Solution
{c}
Count <facial boundary walks> with each edge-side counted once. Their total length is $2m$. Every face walk has length at least four: simplicity excludes loops and parallel-edge two-cycles, absence of triangles excludes length three, and a length-two face walk consisting solely of a bridge traversed twice could be the whole boundary only for the excluded connected graph on two <vertices>. Bridges may occur twice in a longer boundary walk and are still correctly counted. Thus $4f\leq2m$.
The <Euler formula for a connected planar graph> gives $2=n-m+f\leq n-m+m/2=n-m/2$. Rearranging yields
$$
\boxed{m\leq2n-4.}
$$
Back to article page
