probability of chess clubs problem The chess clubs of two schools consists of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that (a) Rebecca and Elise will be paired?
Can someone explain step in step? Because from the book, the solution is
$$\frac{\binom{7}{3}\binom{8}{3}3!}{\binom{8}{4}\binom{9}{4}4!}$$
I really don't know what is going on at all, especially $\dfrac{3!}{4!}$???

There are 24 or (4!)ways to pair the 4 members of the 2 school teams, and 6 or (3!) Ways to pair the members ensuring that rebecca and elise are paired. Hence the probability that rebecca and elise are paired given that they were chosen is 3!/4!. Now we must multiply this probability that both rebecca and elise are chosen to get the probability that rebecca and elise are both chosen and paired together. This means that we multiply $3!/4!$ with $(7C3×8C3)/(8C4×9C4)$, which simplifies to $(7C3×8C3×3!)/(8C4×9C4×4!)$.
