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Worked solution

Treat the buildings as fixed labelled objects

A developer has recently completed a condominium project in HCI Valley. There are 5 blocks of buildings A, B, C, D and E as shown. The developer has 4 colours available. Each block can only be painted using a single colour. Find the number of ways to paint all 5 blocks if

(i) there are no restrictions, mark

(ii) block A and block B cannot be painted with the same colour, marks

(iii) all 4 colours must be used. marks

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Block A is above, B upper right, C lower right, D lower left and E upper left, retaining the source pentagonal placement. These are five labelled fixed buildings, not interchangeable circular seats.
Block A is above, B upper right, C lower right, D lower left and E upper left, retaining the source pentagonal placement. These are five labelled fixed buildings, not interchangeable circular seats.
Block A is above, B upper right, C lower right, D lower left and E upper left, retaining the source pentagonal placement. These are five labelled fixed buildings, not interchangeable circular seats.

[2011/HCI/II/6]

CJC · Review problems · Question 8

Answer

(i) 1024; (ii) 768; (iii) 240.

(i) Independent colours

Each labelled building has 4 choices: 45=10244^5=1024.

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(ii) Method 1: restrict B after choosing A

In A,B,C,D,E order, the numbers of choices are 4,3,4,4,4. Their product is 4×3×43=7684\times3\times4^3=768.

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(ii) Method 2: subtract equal A,B colours

Choose their shared colour in 4 ways and the remaining three building colours in 434^3 ways. Hence 45−4(43)=7684^5-4(4^3)=768.

How did this part go?

(iii) Method 1: choose the repeated-colour pair

Using all four colours on five buildings forces multiplicities 2,1,1,1. Choose the two buildings sharing a colour, then assign the four distinct colours to the pair and three singleton buildings: (52) 4!=240\binom52\,4!=240.

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(iii) Method 2: choose the repeated colour first

Choose which of the four colours repeats and permute the resulting multiset of five colour entries over the buildings: (5!/2!)(41)=240(5!/2!)\binom41=240. Part (iii) is a separate condition and does not inherit the A/B restriction from (ii).

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Common mistakes

The pentagonal placement does not make colourings equivalent under rotation: A–E are fixed labelled blocks.

Solution notes

Source building layout independently reconstructed with all labels. Both printed methods in(ii) and (iii) and every source factor/mark retained.

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