Calculating Colour Combinations for a 4-Square Grid

This problem involves finding the number of ways to colour a 2x2 grid (4 squares) using 3 available colours, with the constraint that no two adjacent squares share the same colour. The top-left square is pre-assigned a colour (Green).

Grid Setup and Constraints

Let the grid be represented as:

A | B
--+--
C | D

• There are 3 colours available (let's denote them as C1, C2, C3).
• The top-left square, A, is fixed as C1 (Green).
• Squares sharing a side are considered adjacent (A is adjacent to B and C; B to A and D; C to A and D; D to B and C).
• Adjacent squares cannot have the same colour.

Step-by-Step Colour Assignment

1. Square A: It is fixed as C1 (Green). This assignment can be done in 1 way.
2. Square B: It is adjacent to A (C1). Therefore, B cannot be C1. B can be C2 or C3. This gives 2 possible colours for B.
3. Square C: It is adjacent to A (C1). Therefore, C cannot be C1. C can also be C2 or C3. This gives 2 possible colours for C.
4. Square D: It is adjacent to both B and C. The number of colour choices for D depends on whether B and C have the same or different colours.

Analyzing Colour Combinations for D

We analyze two main cases based on the colours chosen for squares B and C:

• Case 1: Squares B and C have different colours.
  • If B is C2, then C must be C3 (since C cannot be C1 or B's colour C2).
  • If B is C3, then C must be C2 (since C cannot be C1 or B's colour C3).
  • In this scenario, B and C have different colours (one is C2, the other is C3).
  • Square D is adjacent to B and C (which have different colours). D cannot be the colour of B, nor the colour of C.
  • Therefore, D must take the remaining colour, C1 (Green). There is only 1 choice for D.
  • The number of ways to achieve this case is: (Ways to choose B's colour) $\times$ (Ways to choose C's colour given B's) $\times$ (Ways to choose D's colour) = $2 \times 1 \times 1 = 2$.
• Case 2: Squares B and C have the same colour.
  • B can be C2 or C3 (2 choices). C must be the same as B.
  • Subcase 2a: B = C2 and C = C2.
  • Subcase 2b: B = C3 and C = C3.
  • In this scenario, B and C have the same colour.
  • Square D is adjacent to B and C (which have the same colour). D cannot be the colour used for B and C.
  • If B and C are C2, D cannot be C2. D can be C1 or C3 (2 choices).
  • If B and C are C3, D cannot be C3. D can be C1 or C2 (2 choices).
  • There are 2 ways to choose the common colour for B and C (C2 or C3). For each choice, there are 2 choices for D.
  • The number of ways to achieve this case is: (Ways to choose the common colour for B & C) $\times$ (Ways to choose D's colour) = $2 \times 2 = 4$.

Total Colour Combinations

The total number of possible colour combinations is the sum of the combinations from Case 1 and Case 2.

Total Combinations = (Combinations from Case 1) + (Combinations from Case 2)

Total Combinations = $2 + 4 = 6$.