Divorced Couples Problems Combinatorics 5 couples are sitting in a line but no one wants to sit next to their partner.
How many ways can this be done?
Here's my solution. Is this correct?
$$10!-{5 \choose 1}2! \cdot 9!+{5 \choose 2}2!^2 \cdot 8!-{5 \choose 3}2!^3 \cdot 7!+{5 \choose 4}2!^4 \cdot 6!-{5 \choose 5}2!^5 \cdot 5!$$

Yes, you're converting the current number of couples under consideration into an ordered element and then ordering the elements. Looks fine.
