## 10. Four colors

Each of the fields of the $4 \times 3$ rectangle, as shown in the figure, must be colored with one of four colors. How many ways can this be done so that any four fields sharing a common vertex are colored with different colors?

Result: $\quad 120$

![](https://cdn.mathpix.com/cropped/2024_05_30_3b422eeffac9eaba122eg-7.jpg?height=242&width=315&top_left_y=1359&top_left_x=1550)
## Solution.

Let's denote the colors with $a, b, c$, and fix the color arrangement in the top-left $2 \times 2$ square.

Since 4 different colors can be arranged in these four cells in $4 \cdot 3 \cdot 2 = 24$ ways, we will multiply the number of ways to color the remaining parts of the table by 24 at the end.

![](https://cdn.mathpix.com/cropped/2024_05_30_3b422eeffac9eaba122eg-7.jpg?height=184&width=244&top_left_y=1702&top_left_x=1645)

Let's look at all the ways we can continue coloring to fill the 3x3 square.

The first two rows in the third column must contain the colors $a$ and $c$, and the third row must start with the colors $a$ and $b$.

This leads us to the following three possibilities: