A tennis club has $10$ couples as members. They meet to organize a mixed double match. If each wife refuses to partner as well as oppose her husband in the match, then in how many different ways can the match be arranged?
I made the teams using derangement theorem since making teams in this case is similar to putting $10$ letters in $10$ envelopes such that no letter goes into the correct envelope. So, the number of teams $=\text{D}_{10}=10!\cdot\left(1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\ldots+\dfrac{1}{10!}\right)$
However, I can't figure out how many different matches will be played.
Any help will be appreciated.
Thanks.
As I read it, the question talks about "a" ( = 1) mixed doubles match.
We can choose 2 ladies in ${10\choose 2}= 45$ ways, and since their husbands can't be in the match,
choose the gentlemen in ${8\choose 2} = 28$ ways
Having chosen the 4 who will play, there are only 2 possible pairings.
Thus # of ways the match can be organised = $45\cdot28\cdot2$ = 2520