In how many ways can four married couples sit around a round table so that no man sits next to his own wife?
We denote the couples with the following elements $a_{1} a_{2}, b_{1} b_{2}, c_{1} c_{2}$, and $d_{1} d_{2}$. Since our task is only to ensure that no man sits next to his own wife, we assume that $a_{1}$ always keeps his place. In this case, the other elements can occupy 7! positions. Since $a_{1}$ and $a_{2}$ cannot sit next to each other, $a_{2}$ cannot be placed in the second or the eighth position. Therefore, considering this first condition, the number of different arrangements is

$$
7!-2 \cdot 6!
$$

But $b_{1}$ cannot sit next to $b_{2}$. Therefore, from (1), $2 \cdot 6!$ must be subtracted, but in (1) we have already considered the cases where $b_{1} b_{2}$ sit next to each other and simultaneously $a_{1} a_{2}$ sit next to each other, so $4 \cdot 5!$ must be subtracted from $2 \cdot 6!$. Thus, considering the second condition, from (1) we need to subtract

$$
2 \cdot 6!-4 \cdot 5!
$$

Our third condition is that $c_{1}$ cannot sit next to $c_{2}$. Again, from (1), we need to subtract $2 \cdot 6!$. But in (1) we have already considered the cases where $c_{1} c_{2}$ sit next to each other and simultaneously $a_{1} a_{2}$ sit next to each other, so the subtraction will be $2 \cdot 6!-4 \cdot 5!$. We also need to pay attention to the fact that in (2) we have already considered the cases where $c_{1} c_{2}$ sit next to each other and simultaneously $b_{1} b_{2}$ sit next to each other. Therefore, considering the third condition, from (1) we need to subtract:

$$
(2 \cdot 6!-4 \cdot 5!)-(4 \cdot 5!-8 \cdot 4!)
$$

Considering the fourth condition, that is, $d_{1}$ and $d_{2}$ cannot sit next to each other, from (1) we finally need to subtract:

$$
(2 \cdot 6!-4 \cdot 5!)-(4 \cdot 5!-8 \cdot 4!)+(8 \cdot 4!-16 \cdot 3!)
$$

Therefore, our final result is

$$
\begin{gathered}
7!-2 \cdot 6!-(2 \cdot 6!-4 \cdot 5!)-[(2 \cdot 6!-4 \cdot 5!)-(4 \cdot 5!-8 \cdot 4!)]- \\
-[(2 \cdot 6!-4 \cdot 5!)-2(4 \cdot 5!-8 \cdot 4!)]+(8 \cdot 4!-16 \cdot 3!)
\end{gathered}
$$

Or in another form

$$
\begin{aligned}
7!-2 \cdot 6!- & {[3(2 \cdot 6!-4 \cdot 5!)-12 \cdot 5!+24 \cdot 4!+8 \cdot 4!-16 \cdot 3!]=} \\
& =7!-8 \cdot 6!+24 \cdot 5!-32 \cdot 4!+16 \cdot 3!)]= \\
& =5040-5760+2880-768+96=1488
\end{aligned}
$$

(Károly Rátz, Vienna.)