Maybe, you can calculate number of all ways to place three colors, and then calculate number of ways if you fix one row to be one color and then color other two and subtract from answer. You can to all this with dp.

Lets assume we have the answer for 3*(n-1) grid. Now its obvious that for 3*n grid we just have to take care for the n^th column to be legal.

And the number of ways a column can be legal is 24 ( 3 cells in a column, number of ways to fill is 27 and number of illegal ways would be 3 i.e RRR, GGG, BBB ).

Lets assume we call answer for 3*(n-1) to be dp[n-1] then we can write it as:
dp[n] = dp[n-1]*24 + something
This plus something is because some of the illegal ways for coloring 3*(n-1) can also be made legal.
Now its obvious that these illegal ways for (n-1) would be only illegal in terms of rows and not columns.
Lets see how many illegal ways of rows exists!
Either one of the rows can be illegal or 2 rows can be illegal. Note that all 3 rows cannot be simultaneously be illegal.
And hence number of one row illegal = ³C₁*3
Similarly for 2 rows = ³C₂*3
And considering the coloring options for these 2 conditions, we get:

dp[n] = dp[n-1]*24 + Σ ³C_i*3*2ⁱ for all i ∈ {1,2}


P.S — I'm obviously unsure of the solution and would like to test it. So please provide a link to the problem.

$$$EDIT: $$$ Its wrong the ideia, sorry.

As i_love_cuiaoxiang mentioned, there is 24 choices of column to respect the column restriction.

Lets do a dp of coloring choices that respect the column restriction and not respect the row restriction.

The $$$dp[i][j]$$$ have the number of choices that don't respect the row restriction if we just consider the first $$$i$$$ columns and put $$$j$$$ in the $$$i$$$ column ($$$j$$$ can be the options that we can put in the column (24), can be "RRG", "BBR" or any other combination except "RRR", "GGG" and "BBB").

How the transition works? $$$dp[i][j] = \sum\limits_{k} dp[i-1][k]$$$ if $$$k$$$ have a common value with j in any position("RGB" have a common value with "RRB" but don't have with "GBR", note that position matters).

The answer for the problem will be $$$24^{n} - \sum\limits_{i} dp[n][k]$$$. The complexity will be $$$O(n*24^2)$$$.

I think my solution would work if we already precompute answers for n=1 and n=2!

I suggest this solution:

Let:

• $$$sum =$$$ number of ways to create a table that no column have all the cells of same color.
• $$$sum1 =$$$ number of ways to create a table that no column have all the cells of same color and have one or more rows have all the cells of same color.

-> $$$answer = sum - sum1$$$

$$$sum$$$ & $$$sum1$$$ can be calculated by dp bitmask with Inclusion — Exclusion.