Painting Squares

For each colour that occupies the diagonal line of a 3x3 grid, there are 4 unique configurations, all of which can be achieved by rotating the grid through consecutive 90 degrees turns.

Once a full rotation has been completed, switch the colour and repeat until all 3 colours have occupied the diagonal line and completed full rotations.

$3 \times 4 = \boxed{12}$

It's not as mathematically pleasing (or adaptable to larger grids), but still solves this particular problem.

If we number the squares as follows:

There are six ways to paint 1 and 2.

Then, given 1 and 2, there are two ways to paint 4.

After that, there is only one way to paint the remaining boxes.

Therefore, there are $6 \cdot 2 = \boxed{12}$ ways to paint the squares.

Relevant wiki: Derangements

Consider the first row, the number of permutations for 3 colors is $3!=6$ . When the first row is fixed, the number of permutations in the second row is the derangement of 3, meaning the number of permutations such that they are not in their original position, denote as $!3=2$ . The last row is forced by the first two rows. Hence the answer is $6\times 2=12$ .

Once you paint marked squares (see picture), there is just one way to paint the rest 4 of them in order to get 3x3 grid with three colors such that each row and column has all three distinct colors. Hence, we should only determine how many ways are there to paint 5 marked squares.

There are $3\times 2= 6$ ways to paint first-column squares. Then, there are two ways of painting the left two squares knowing colors of first-column squares. Hence, there are $6\times 2= 12$ ways to paint 5 marked squares.

Thus, answer is 12.

Let the colors are $c_1,c_2,c_3$ .

Let $S$ be set of all permutations of the above three colors.

So, $S=\{(c_1,c_2,c_3);(c_1,c_3,c_2) ; (c_2,c_1,c_3);(c_2,c_3,c_1);(c_3,c_2,c_1);(c_3,c_1,c_2)\}$

As per the above requirement the coloring can be done by using three distinct element from $S$ such that position of colors in every element is different.So, here there will two such subsets these are

$S_1=\{(c_1,c_2,c_3);(c_2,c_3,c_1);(c_3,c_1,c_2)\}$ and

$S_2=\{(c_1,c_3,c_2);(c_2,c_1,c_3);(c_3,c_2,c_1)\}$

From $S_!$ we can add these elements in the rows (columns) and they can permute along rows.So there will be $3!=6$ ways.

Similarly from $S_2$ there are also $6$ ways.

So, total ways $=12$

If we consider the center square then there is 3 choices​ to colour that and for the 4 squares​ at it's 4 sides, two of different rows have 2 choices​ and others have only 1 choice.so total number of ways to make 3×3 colour cube = 3ç¹×2ç¹×2ç¹=3×2×2=12

This is probability question, where number of angles are 4 (90, 180, 270, 360/0) Placing each color diagonally can be done in 3 ways

So total number of ways of arranging it is 4c3= (4 * 3 * 2) / (2 * 1) = 12

Let's consider the three colors as permutation-based objects. In the first row, the colors can be arranged in 3! = 6 ways - since it is the first row, the colors within the first row could be arranged in any way. In the second row, the colors can be arranged in a very limited way, since we have some of the 6 ways having a same color on the column or row. So, the colors within the second row could be arranged by (3 - 1)! = 2 ways. In the third row, the colors wouldn't be able to arrange themselves freely, since we're preventing the colors from other columns or rows to be similar. So there are only (3 -2)! ways = 1 way to arrange the colors in the third row. Therefore, we have 6 x 2 x 1 = 12 ways to distinctively color the 3 x 3 square without having a same color in either row or column.

For a 3 by 3 grid it is quite easy to calculate the number of ways when 3 colours are used. Each colour can be used once on the diagonal, with the other two colours each occupying a corner with the two adjacent squares filled with the other colour. (Think also of chess and the move of a knight.) This yields 3 ways. Each of these 3 ways are rotated: 0 degrees, 90 degrees, 180 degrees and 270 degrees. This gives another 4 ways for each of the 3 colours. 3 x 4 = 12 ways in total.

hint : this have something to do with derangement , would be more challenging for 4*4 or more .

Isn't this just the number of Latin squares of order 3?

As the problem is stated, 18 is the correct answer. The problem asks about each row or column, not and column. For each row or column, there are 3! = 6 arrangements, and there are 3 rows (or columns).