11.2. Find the number of different ways to arrange all natural numbers from 1 to 9 inclusive in the cells of a 3 by 3 table, one number per cell, such that the sums of the numbers in each row and each column are equal. The table cannot be rotated or reflected.
Answer: 72 ways

Solution. Among the numbers from 1 to 9, there are 5 odd numbers. Since the sums in all rows and columns are $\frac{1}{3}(1+2+\ldots+9)=15$ - which are odd, there must be an odd number of odd numbers in each row and each column. This is only possible if one row contains three odd numbers, and the other rows contain one each, and one column contains three odd numbers, and the other columns contain one each. Therefore, all odd numbers must be in the union of a certain row and a certain column. At the intersection of these row and column, the number $15+15-(1+3+5+7+9)=5$ is written. Moreover, 7 and 9 cannot be in the same row or column. Therefore, we can write the number 5 in one of the nine cells, then write the number 9 in one of the four cells of the same row or column as 5, and then write the number 7 in two cells of the same row or column, not adjacent to 9. This gives us 9.4$\cdot$2=72 ways. Then we verify that at the intersection of the column and row containing 7 and 9, only 2 can be placed: 6 and 8 do not fit the sum, and 4 would give another 4 with 7. After this, it is clear that the remaining even numbers can be placed in the remaining cells of the table, and there is only one way to do this.

Grading Criteria. Noted that all odd numbers are in the union of a certain row and a certain column: 2 points.

Proved that the number 5 is written at the intersection of these row and column: 1 point.

Noted that 7 and 9 cannot be in the same row or column: 1 point.

Arrangements of the remaining odd digits in 72 ways: 2 points.

Lack of precise justification that the sums in rows and columns are 15: deduct 1 point. Arrangements differing by rotation or reflection are considered the same: deduct 2 points.

Loss of part of the solutions: deduct 2 points or more.

All required arrangements are constructed, but it is not proved that there are no others: no more than 3 points.