Three couples dine at the same restaurant every Saturday at the same round table. The restaurant's policy is:

(a) never to seat husband and wife as neighbors at the table;

(b) the arrangement of the six at the table is different each Saturday.

For how many Saturdays can the couples go to the restaurant without repeating the seating arrangements?
To simplify, let's denote each couple by a pair of numbers: $(1,2),(3,4),(5,6)$, where in each pair, one number represents the husband and the other the wife. Three pairs cannot be neighbors: $(1,2),(3,4)$, $(5,6)$.

See two possible arrangements; starting from 1 in a clockwise direction: 1-3-2-5-4-6 and 1-6-4-5-2-3.

By fixing the position of 1 at the table and reading the numbers formed in a clockwise direction, the problem reduces to finding all 6-digit numbers that can be written with the digits $1,2,3,4,5$ and 6, where:

- all numbers start with the digit 1;
- 1 and 2, 3 and 4, 5 and 6 cannot appear together.

We find the 16 numbers listed in the table.

| 132546 | 132645 | 135246 | 135264 | 135426 | 136245 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 136254 | 136425 | 142536 | 142635 | 145236 | 145326 |
| 146235 | 146325 | 153246 | 154236 |  |  |

Therefore, the answer is 16 Saturdays.