Flipping 3-Coin Triangles

Consider the `coloring' of the coins shown on the right (R = red, G = green, B = blue). Count how many heads there are for each color. In the initial state, the counts are: $R, G, B = 1, 2, 2$ .

Each move involves one coin of each color. Hence, the count for each color changes by 1. Consequently, the parity (even/oddness) of the differences between pairs of colors is invariant (does not change).

The counts for the desired final state are: $R, G, B = 3, 4, 3$ . Here are all the counts and the parities of the pairwise differences: \begin{array}{l|*{3}{c|}*{3}{|c}} \text{state} & R & G & B & R - G & G - B & B - R \\\hline \text{initial} & 1 & 2 & 2 & \text{odd} & \text{even} & \text{odd} \\ \text{final} & 3 & 4 & 3 & \text{odd} & \text{odd} & \text{even} \end{array} Since the parities of the final state differ from those of the initial state, it is impossible to transform the initial state to the final state with these moves.

Consider a colouring arrangement of the ten coins (red, blue and black) as shown above. Note that a flip of 3 coins will change the state of a single coin of each colour.

We wish to transform the tails into heads. Therefore, we need to change the state of 1 red, 2 blues and 2 blacks. These numbers clearly aren't possible due to divisibility by 3.

The answer is that it was impossible because they can have an random orangement