= Solution
{c}
A triangle $x_1y_1y_2$ forces at least three vertex colours. Three suffice: colour $x_1,x_2$ blue; $y_1,y_3,y_4$ orange; and $y_2,y_5$ green, as in the drawing. Hence \b[the <chromatic number> is three].
The five <edges> incident to $x_1$ force at least five edge colours. An explicit proper five-edge-colouring is
$$
\begin{array}{c|ccccc}
&y_1&y_2&y_3&y_4&y_5\\\hline
x_1y_j&1&2&3&4&5\\
x_2y_j&2&3&4&5&1
\end{array}
$$
Give $y_1y_2$ colour 5 and $y_1y_5$ colour 3. At $y_1$ the incident colours are $1,2,5,3$; at $y_2$ they are $2,3,5$; at $y_5$ they are $5,1,3$; all are distinct. The two remaining <vertices> each have two distinct incident colours. Thus \b[the <edge chromatic number> is five].
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