Number of colourings of the vertices of a square

Question

$(1)$ The vertices $A,B,C,D$ of a square are to be coloured with one of three colours red, blue, or green such that adjacent vertices get different colours. What is the number of such colourings?

What I attempted :-

The vertex $A$ can be coloured with $3$ colours in $3$ ways. There are $2$ choices for $B$ and $C$ (One restriction for each)
There is $1$ choice for $D$ (Two restrictions)

Required number of colourings = $3\times 2 \times 2 \times 1=12$

Am I Correct ?

Answer 1

It could be that $C$ and $A$ get the same color, which would leave two choices for $D$. We need to condition on that. There is one choice for $C$ that is the same as $A$ and one different, so it becomes $3 \times 2 \times 1 \times 2 + 3 \times 2 \times 1 \times 1=18$

Answer 2

Possibilities:

1. A has free choice, B and C are the same color, and D is either the same color as A or the last remaining color. There are 3 ways to choose the color for A, 2 ways to choose for B, 1 way to choose for C (since it is the same as B), and 2 ways to choose for D. This gives 12 possible colorings.

2. A has a free choice, B and C are different colors, and D must be the same color as A. Then you have 3 choices for A, 2 for B, and 1 for each of C and D. There are 6 of these colorings.

   Those are the only possibilities. So, there are a total of 18 colorings.