8.1. (12 points) In how many ways can eight of the nine digits $1,2,3,4,5,6$, 7,8 and 9 be placed in a $4 \times 2$ table (4 rows, 2 columns) so that the sum of the digits in each row, starting from the second, is 1 more than in the previous one?
# Answer: 64.

Solution. The sum of all nine numbers is 45. Let $x$ be the sum of the two numbers in the first row, and let $a$ be the one number out of the nine that we do not place in the figure. Then $x+x+1+x+2+x+3=45-a$, from which $4 x+a=39$. Since $a$ is an integer from 1 to 9, we get 2 possible cases: either $x=9, a=3$, or $x=8, a=7$.

If $a=3$, then we need to place the numbers $1,2,4,5,6,7,8,9$, and the sum of the numbers in the rows should be 9, 10, 11, and 12, respectively. Possible variants for the numbers in the first row: $9=1+8=2+7=4+5$.

If the first row contains 1 and 8, then the second row should contain 4 and 6, the third row 2 and 9, and the last row 5 and 7.

If the first row contains 2 and 7, then the second row should contain 1 and 9 (with other variants, it is impossible to select numbers for the third row), the third row 5 and 6, and the last row 4 and 8.

If the first row contains 4 and 5, then the second row can have 1 and 9 or 2 and 8. But in both cases, it is impossible to find numbers for the third row.

If $a=7$, then we need to place the numbers $1,2,3,4,5,6,8,9$, and the sum of the numbers in the rows should be 8, 9, 10, and 11, respectively. The sum of the numbers in the first row is $8=5+3=6+2$.

If the first row contains 3 and 5, then the second row should contain 1 and 8, the third row 6 and 4, and the last row 2 and 9.

If the first row contains 6 and 2, then the second row can contain 1 and 8 or 4 and 5. If they contain 1 and 8, it is impossible to select numbers for the next row. Therefore, they contain 4 and 5, then the next row contains 1 and 9, and the last row 3 and 8.

Thus, we have 4 variants of number placement without considering the order of numbers in the rows. Since the numbers in each row can be swapped, we get $2^{4}=16$ variants for each placement. In total, we get 64 variants.