13. In a $4 \times 4$ chessboard, 8 small squares are to be colored black such that each row and each column has exactly two black squares. Then there are $\qquad$ different ways to do this (answer with a number).
13. 90 .

The first row can be colored with 2 black cells in $\mathrm{C}_{4}^{2}$ ways. After the first row is colored, there are the following three scenarios:
(1) The black cells in the second row are in the same columns as those in the first row, in which case, the remaining rows have only one way to be colored;
(2) The black cells in the second row are in different columns from those in the first row, in which case, the third row has $\mathrm{C}_{4}^{2}$ ways to be colored, and the coloring of the fourth row is then determined;
(3) The second row has exactly one black cell in the same column as one in the first row, and there are 4 such coloring methods. After the first and second rows are colored, the third row must have 1 black cell in a column different from the ones above, in which case, the third row has 2 ways to be colored, and the coloring of the fourth row is then determined.
Therefore, the total number of different coloring methods is
$$
6 \times(1+6+4 \times 2)=90 \text {. }
$$