A rectangular wall is partitioned into 5 rectangular parts as shown in the following diagram. If you have three different colours to paint the wall and if no two adjacent parts are to be painted with the same colour, in how many different ways can one paint the wall?
18
Label the five parts as shown in the diagram: top-left (TL), bottom-left (BL), top-middle (TM), bottom-middle (BM), and the single full-height part on the right (R).
From the diagram, the pairs that share a boundary (and so must get different colours) are: TL–BL, TL–TM, BL–BM, TM–BM, TM–R, and BM–R.
TM, BM, and R are mutually adjacent to each other (TM–BM, TM–R, and BM–R are all shared edges), so all three must get three different colours. With 3 colours available, this can be done in 3! = 6 ways.
For each such choice, TL must differ from TM (2 remaining colours) and BL must differ from BM (2 remaining colours), giving 4 combinations for (TL, BL); but TL and BL are themselves adjacent, so the 1 combination where they would coincide (both equal to the one colour used by neither TM nor BM) is invalid, leaving 4 − 1 = 3 valid combinations.
Multiplying, the total number of ways is 6 × 3 = 18 based on the adjacency shown in the diagram.
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