# Answer: 14.

Solution. The mine adjacent to the number 1 is in the second, third, or fourth column.

1 case. If the mine is in the second column, it has three possible positions, and it is adjacent to the number two in the first column, which means the second mine adjacent to this two is in the first column, and it has two possible positions. In this case, we get $2 \cdot 3=6$ possible mine placements.

2 case. If the mine is in the fourth column, due to symmetry, we also get 6 possible placements.

3 case. If the mine is in the third column, it has two possible positions in this column. The mines adjacent to the two must be in the same column as the two. In this case, we get two possible placements.

Summing up the results, we get a total of $6+6+2=14$.

Criteria. All cases of mine placement are provided, but it is not shown that there are no others: 4 points.
4. Some squares of the table are mined. Each number written in a square shows the number of mines in the squares adjacent to the given square. (See the figure. Adjacent are squares that have at least one common point; a square with a number is not mined). In how many ways can the mines be placed in the table? Justify your answer.

![](https://cdn.mathpix.com/cropped/2024_05_06_de7f3f58c4a58f7306afg-1.jpg?height=228&width=344&top_left_y=289&top_left_x=2379)
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