A rectangular wall is divided into four squares of equal size where there are two rows each having two squares. The top-left square is coloured with green. If, including green, there are three colours available and each square is coloured using any one of these three colours such that no two adjacent squares get painted with the same colour, then how many colour combinations are possible?
- (a)2
- (b)4
- (c)6
- (d)8
Correct — C, 6. In a two-by-two grid only edge-sharing squares are adjacent; the two diagonal squares touch at a corner alone and may repeat a colour. With the top left fixed as green, the top right has 2 admissible colours and the bottom left has 2, giving four cases. In the two cases where those two squares carry the same colour, the bottom right avoids only that one colour and has 2 choices. In the two cases where they differ, the bottom right must avoid both and has exactly 1. So the count is 2 × 2 + 2 × 1 = 6.
- (a)2 — Counts only the arrangements in which the diagonal pairs repeat, that is green facing green and the other colour facing itself. Those are two of the six, not all of them.
- (b)4 — Takes 2 × 2 for the two neighbours of green and then assumes the last square is always forced. It is forced only in the two cases where its two neighbours already differ.
- (d)8 — Multiplies 2 × 2 × 2, giving the last square two free choices every time. In the two cases where its neighbours carry different colours only one colour is left to it, so this over-counts by two.
Colouring problems of this kind are counted by fixing one cell and then walking the adjacency structure, splitting into cases wherever the number of remaining choices changes. The four squares of a two-by-two wall form a closed loop of adjacencies — top left to top right to bottom right to bottom left and back — because the diagonal pairs are not adjacent. It is that loop, not the grid picture, that governs the count.
The whole item turns on what "adjacent" means. If the diagonals counted as adjacent, all four squares would be mutually adjacent, three colours could not be enough, and the answer would be zero. Reading adjacency as edge-sharing gives the loop, and the loop is what makes the case split necessary: the fourth square is sometimes free and sometimes forced. Do not try to enumerate all 3⁴ = 81 colourings under exam pressure. Fixing the green corner and walking round the loop takes four lines. There is also a formula worth carrying — the number of proper colourings of a loop of n cells with k colours is (k − 1)ⁿ + (−1)ⁿ(k − 1) — which for n = 4 and k = 3 gives 16 + 2 = 18, and dividing by 3 to account for the fixed green corner returns 6.
- Adjacent means sharing an edge; the diagonal squares of a two-by-two grid share only a corner and may repeat a colour.
- The adjacency structure of a two-by-two grid is a closed loop of four cells.
- With the first square fixed, each of its two neighbours has k − 1 = 2 admissible colours.
- The fourth square has 3 minus the number of distinct colours on its two neighbours, so 2 in one case and 1 in the other.
- Proper colourings of a loop of n cells with k colours number (k − 1)ⁿ + (−1)ⁿ(k − 1); for n = 4, k = 3 that is 18, and 18 ÷ 3 = 6 once one cell's colour is fixed.
The diagonal squares are not adjacent, which is what leaves the bottom right free in the first case.
- Treating the diagonal squares as adjacent, which makes the task impossible with three colours.
- Multiplying 2 × 2 × 2 as if the last square always had two options.
- Forgetting that green itself is available again for the squares that do not touch the green one.
Asked as a small colouring puzzle with one cell pre-set, testing whether you split the count into cases instead of multiplying blindly.
In how many ways can four children be made to stand in a line such that two of them, A and B, are always together?
- (a) 6
- (b) 12
- (c) 18
- (d) 24
Answer(b) 12
The same counting discipline applied to a restriction instead of a colour rule. Both items are solved by fixing what the condition forces first and multiplying the free choices afterwards, rather than listing every arrangement.
- practice — not a real PYQ
The four squares of a two-by-two grid are to be coloured from four available colours so that no two squares sharing an edge get the same colour. If the top left square is fixed as red, how many colourings are possible?
- (a)12
- (b)16
- (c)21
- (d)24
Answer(c) 21 — each neighbour of red has 3 choices; in the 3 cases where those two agree the last square has 3 choices, and in the 6 cases where they differ it has 2, so 9 + 12 = 21.
- practice — not a real PYQ
Three squares are placed in a row. Each is to be coloured with one of three colours so that no two squares next to each other share a colour. How many colourings are possible?
- (a)6
- (b)9
- (c)12
- (d)27
Answer(c) 12 — the first square has 3 choices and each later one has 2, so 3 × 2 × 2 = 12.