Pairing a chess match

Sample space is the set of all possible pairs between both clubs. There are $\binom{8}{4}$ waysof choosing the first team, for wich there are $\binom{9}{4}$ ways of chosing the second team, for wich there are $4!$ ways to pair once both teams have been chosen, so sample space has a cardinal of $\binom{8}{4}\binom{9}{4}*4!$. The pairing is done by fixing a group and making to permute all mebers from the other group.

a)There are $\binom{7}{3}$ possible teams having Rebecca as a member, for wich there are $\binom{8}{3}$ possible teams having Elise as a memeber, for wich there are $3!$ possible pairs given that Rebecca and Elisa are paired. Therefore there are $3!\binom{7}{3}\binom{8}{3}$ ways in wich Rebecca and Elise can be paired.

b)Considering incise a), from $4!$ possible pairs, there are $4!-3!=18$ pairs where Rebecca and Elise are not paired after choosing the teams, so the new event of interest has a cardinal of $18\binom{7}{3}\binom{8}{3}$.

You are making this harder than it is. In a), the first stage, i.e. how many are selected to the teams, is a red herring. All possible matches ($8\ast 9=72$) are equally likely. Hence the answer to $a$ is simply $1/72$.

In b) there is a $4/8=1/2$ chance that Rebecca will be chosen and a $4/9$ chance that Elise will be shosen. These events are independent so the probability both will be chosen is $\frac{1}{2}\frac{4}{9}=\frac{2}{9}$. Then simply subtract the probability in a) to get the answer, i.e. the answer is $\frac{2}{9}-\frac{1}{72}=\frac{15}{72}$.

c) is simply $1$ minus the chance none of them will be chosen.