Example 55 A number of couples sit in a row. Find the number of arrangements in which no couple sits next to each other.

Keep the original text's line breaks and format, and output the translation result directly.
Let $S$ be the set of all possible arrangements of 5 couples in a row, and let $A_{i}$ be the set of arrangements where the $i$-th couple is adjacent. Therefore, by the principle of inclusion-exclusion, the number of arrangements where no couple is adjacent is
$$
\begin{array}{l}
\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3} \cap \bar{A}_{4} \cap \bar{A}_{5}\right|=|S|-\sum_{i=1}^{5}\left|A_{i}\right|+\sum_{1 \leqslant i<j \leqslant 5}\left|A_{i} \cap A_{j}\right|- \\
\sum_{1 \leqslant i<j<k \leqslant 5}\left|A_{i} \cap A_{j} \cap A_{k}\right|+\sum_{1 \leqslant i<j<k<k \leqslant 5}\left|A_{i} \cap A_{j} \cap A_{k} \cap A_{t}\right|, \\
\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{1} \cap A_{5}\right| \\
=10!-C_{5}^{1} \times 2 \times 9!+C_{5}^{2} \times 2^{2} \times 8!-C_{5}^{3} \times 2^{3} \times 7!+C_{5}^{4} \times 2^{4} \times 6!-C_{5}^{5} \times 2^{5} \times 5! \\
=3628800-10 \times 362880+40 \times 40320-80 \times 5040+80 \times 720-32 \times 120 \\
=1363360 . \\
\end{array}
$$