Tricky Tennis Twins

Let $A$ and $a$ refer to the first set of twin sisters, $B$ and $b$ refer to the second set of twin sisters, and $C$ and $c$ refer to the third set of twins. Similarly, let $1$ and $2$ refer to the husbands of $A$ and $a$ respectively, $3$ and $4$ refer to the husbands of $B$ and $b$ , and $5$ and $6$ refer to the husbands of $C$ and $c$ .

From the conditions of the problem, there are 4 possible men to be in a team with each of the women. Since there are 6 women, this means a total of 24 possible teams are possible. These are

$\begin{array}{cccccc} A3 & a3 & B1 & b1 & C1 & c1\\ A4 & a4 & B2 & b2 & C2 & c2\\ A5 & a5 & B5 & b5 & C3 & c3\\ A6 & a6 & B6 & b6 & C4 & c4\\ \end{array}$

We will now determine the number of ways to have 6 teams to create this playoff. Firstly, notice that if $FM$ and $fm$ are teams, where $F$ and $f$ are twin sisters, $Fm$ and $fM$ are also teams. Since there are 3 pairs of this in each rally, we will have 8 combinations for each rally of the form $F_1 M_1$ , $f_1, m_1$ , $F_2 M_2$ , $f_2, m_2$ , $F_3 M_3$ and $f_3, m_3$ .

Case 1: A set of twins is paired with another set of twins.

If two of the teams are $A3$ and $a4$ (or vice-versa), then twins $C$ and $c$ are left with $1$ and $2$ , which results in twins $B$ and $b$ left with $5$ and $6$ .

If two of the teams are $A5$ and $a6$ (or vice-versa), then twins $B$ and $b$ are left with $1$ and $2$ , which results in twins $C$ and $c$ left with $5$ and $6$ .

Over all these cases, there are 2 distinct sets of playoff teams possible (that is, of the form $F_1 M_1$ , $f_1, m_1$ , $F_2 M_2$ , $f_2, m_2$ , $F_3 M_3$ and $f_3, m_3$ ).

Case 2: A set of twins is not paired with another set of twins

If two of the teams are $A3$ and $a5$ (or vice versa), then twins $B$ and $b$ must have one of their parters as $6$ (since $C$ and $c$ cannot be paired with $6$ if $B$ and $b$ are with $1$ and $2$ ). However, $B$ and $b$ can have either of $1$ and $2$ as their other partner. This results in $C$ and $c$ with $4$ and the either of $1$ and $2$ that has not been chosen already. Thus, there are 2 distinct sets of playoff teams possible.

If two of the teams are $A3$ and $a6$ (or vice-versa), similarly there are 2 distinct sets of playoff teams.

If two of the teams are $A4$ and $a5$ (or vice-versa), similarly there are 2 distinct sets of playoff teams.

If two of the teams are $A4$ and $a6$ (or vice-versa), similarly there are 2 distinct sets of playoff teams.

Overall, there are 10 possible distinct sets of playoff teams, each with 8 combinations. Thus, there are 80 possible playoff teams.