A quadrilateral has four vertices $$𝐴, 𝐵, 𝐶, 𝐷$$. We want to colour each vertex in one of the four colours red, blue, green or yellow, so that every side of the quadrilateral and the diagonal $$𝐴𝐶$$ have end points of different colours. In how many ways can we do this?

Let the four vertices be labelled in order as $$A,B,C,D$$ so that the sides are $$AB,\,BC,\,CD,\,DA$$ and the extra restriction is on the diagonal $$AC$$. A proper colouring demands that the two endpoints of every one of these five edges receive different colours.

Step 1 - Choose a colour for $$A$$. Any of the four colours (red, blue, green, yellow) may be used.
    Number of choices for $$A = 4$$.

Step 2 - Colour $$B$$. Edge $$AB$$ forces $$B$$ to be different from $$A$$, leaving three colours.
    Number of choices for $$B = 3$$.

Step 3 - Colour $$C$$. Vertex $$C$$ is adjacent to both $$B$$ (edge $$BC$$) and $$A$$ (diagonal $$AC$$), so its colour must differ from the colours on $$A$$ and $$B$$. The colours on $$A$$ and $$B$$ are already different, so exactly two colours remain available.
    Number of choices for $$C = 4-2 = 2$$.

Step 4 - Colour $$D$$. Vertex $$D$$ is adjacent to $$C$$ (edge $$CD$$) and to $$A$$ (edge $$DA$$). Hence $$D$$ must avoid the colours used on $$A$$ and $$C$$. Since those two colours are distinct, again two colours remain.
    Number of choices for $$D = 4-2 = 2$$.

Total colourings = $$4 \times 3 \times 2 \times 2 = 48$$.

Thus the number of ways to colour the quadrilateral so that every side and the diagonal $$AC$$ have differently coloured endpoints is 48.