Condition (a) requires that each of $R$, $Y$, $B$ is to occur exactly twice in each row and column.

We apply (b) as follows: Take any unit square of a given colour, say $R$. As explained above, its row and column must each contain another $R$. The three $R$'s are the vertices of a rectangle and the fourth vertex must also be $R$ by condition (b).

If (a) and (b) hold for a given matrix, then they will also hold after any permutation of the rows and columns of the matrix.

Let $N$ denote the number of Colombian Squares for which the 1st row and column in the table are both coloured in the pattern $RYBRYB$. There are $\binom{6}{2} \binom{4}{2} \binom{2}{2} = 90$ ways to colour the first row, and for each of these choices there are $\binom{5}{2} \binom{3}{2} = 30$ ways to colour the remainder of the first column, so the total number of matrices will be $90 \cdot 30 \cdot N = 2700N$.

To find $N$, we examine those Colombian Squares which partially look like

When we attempt to fill in the 4th row and column with $Y$, $B$, with the help of (b) we quickly dismiss all but the following four patterns:

(i)

(ii)

(iii)

(iv)

Note that (iv) is obtained from (iii) by swapping rows with columns.

In each of the cases above, once we fill in the first empty $2 \times 2$ table, the other entries are determined by (a) and (b), applied to the $Y$'s and $B$'s.

In case (i), in columns 2 and 5 the B can only be in rows 2 and 5, because of (a). Similarly, the Y in columns 3 and 6 can only be in rows 3 and 6. The unfilled unit squares are then to be filled with R, using (a) again. Therefore, the only possibility in this case is this:

(i)

In case (ii), by using (b), we see that the distribution of Ys and Bs surrounding the top left empty $2 \times 2$ cell forces all its four entries to be R. In turn, this implies that all the entries in the top right empty $2 \times 2$ cell are Y or B. But, because of condition (a), this is impossible as there are already a Y and a B entry in these columns. Therefore, case (ii) is impossible.

In case (iii), proceeding as before, we are left with four possible Colombian Squares:

(iii)

In case (iv) the four possibilities are obtained from those in case (iii) by swapping rows with columns. Hence, we get $N = 1 + 4 + 4 = 9$ possibilities.

Therefore, there are $9 \times 2700 = 24300$ Colombian Squares.

Instead of fixing a row and a column, we first focus on one colour. Let us see in how many ways we can arrange the yellow squares in the table such that there are exactly two yellow squares in each row/column (this is condition a). By condition (b), the 12 yellow squares in the table form the vertices of 3 different rectangles.

There are $\binom{6}{2} = 15$ ways to choose two yellow squares in column 1. There are 5 ways to choose another column, so that the yellow squares in the 1st column and the newly chosen column form the vertices of a rectangle. There are thus 75 ways to choose the first rectangle with yellow vertices.

For the second rectangle, we choose the next available column (either the 2nd or if taken, the 3rd). There are $\binom{4}{2} = 6$ ways to choose yellow squares in it, so that they are not in the same rows as the yellow unit squares of the first rectangle. Then, there are 3 ways to choose a new column and thus complete the second rectangle with yellow vertices. This gives 18 possibilities for the 2nd rectangle. The third rectangle will then be uniquely determined by condition (a). Thus in total, there are $75 \times 18 = 1350$ ways of arranging the yellow squares.

Let us first determine how we can fill in the remaining squares in the first column, and then, for each case, how many Colombian tables we get.

The first 2 squares in the 3rd row could be RR, BB, RB or BR. This gives four cases to consider.

(i)

(ii)

(iii)

(iv)

In the first case, there are 3 options for completing this RR to a rectangle, and then all the colours in the table are determined by condition (a). Similarly in case (ii), we have 3 options for filling in the other BB in the first two columns, then everything else is determined.

In case (iii) columns 1 and 2 cannot be RR or BB in any row, because of condition (b). Therefore, and because of condition (a), there is another row starting with RB and there are 3 possibilities to choose it. For each such choice, there are two ways to complete the two rows starting with $RB$. Indeed, the two free squares left in these rows can be filled in with $RB$, or alternatively with $BR$. The colours of other squares in the table are then completely determined by condition (a). This leads to 6 ways of completing the table to a Colombian square in case (iii). Exchanging $R$ and $B$ switches from case (iii) to case (iv), hence there are 6 Colombian squares in case (iv) as well.

In total, cases (i)–(iv) yield $3+3+6+6 = 18$ possibilities. Hence, the number of Colombian squares is equal to $1350 \times 18 = 24,300$.