10,11 |  |
| $[\underline{\text { Rule of Product }}]$ |  |  |

Given the table (see figure).

| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 7 | 1 | 2 | 3 | 4 | 5 | 6 |
| 6 | 7 | 1 | 2 | 3 | 4 | 5 |
| 5 | 6 | 7 | 1 | 2 | 3 | 4 |
| 4 | 5 | 6 | 7 | 1 | 2 | 3 |
| 3 | 4 | 5 | 6 | 7 | 1 | 2 |
| 2 | 3 | 4 | 5 | 6 | 7 | 1 |

You can swap rows and columns in any order.

How many different tables can be obtained this way from the given table?
Note that the numbers in each row of the table are distinct. Moreover, for any rectangular cell, the sum of the numbers in its opposite corners is equal to the sum of the numbers in the other two opposite corners modulo 7. These properties are obviously preserved under the allowed permutations. Therefore, the final table is completely determined by its top row and left column. On the other hand, by permuting the columns, we can obtain any of the 7! permutations of the numbers in the top row, and by permuting the rows from the second to the seventh, we can obtain any of the 6! permutations of the numbers in the six lower cells of the left column.

## Answer

$7!\cdot 6!$ tables.