The chess clubs of two schools consist 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
Rebecca and Elise will be paired?
I believe it's ((8 Choose 4)*(4 Choose 1))X((9 Choose 4)*(4 Choose 1))/((17 Choose 8)*(8 Choose 2))
Introduction:
The first school has 8 players in their chess club, out of which, 4 are to be selected.
The first school has 9 players in their chess club, out of which, 4 are to be selected.
Each selected player from the first school (out of the 4 selected players from the first school), is paired randomly to play against a selected player from the second school (out of the 4 selected players from the second school).
The added constraint is that, Rebecca and her sister Elise belong to the chess clubs of the two different schools.
In order for Rebecca and Elise to be paired, each of them must first be selected from their respective schools, among the 4 members chosen for the contest. After that, the pairing of players between the two schools must be such that Rebecca and Elise must be selected in one pair, where they play against each other.
Total number of possible cases:
Without any added constraint, 4 players can be selected for the contest from the 8 players of the first school chess club in _{8}C_{4} ways.
Similarly, and independent of the selection of the first school, 4 players can be selected for the contest from the 9 players of the second school chess club in _{9}C_{4} ways.
After 4 players of each school are selected, the pairing is done as follows:
Thus, the pairing of the players from the two schools, in order to get each pair with one player from the first school and one player from the second school to play against each other, can be done in 4! ways.
Thus, the total number of ways (without any constraints), in which 4 players are selected from each of the two schools, and each player from one school is paired to play against one player of the other school, is 4! (_{8}C_{4}) (_{9}C_{4}).
Number of favorable cases:
Now, Rebecca and Elise go to two different schools and each one is selected from their respective schools. So, after already choosing the 2 sisters, there remain 7 (= 8 – 1) members in the first school chess club, and 8 (= 9 – 1) members in the second school chess club.
As 1 player is already chosen from each school for the contest (out of the 4 required players), from each of the 2 schools, 3 (= 4 – 1) more must be chosen, for each school.
For the first school, 3 more players can be chosen for the contest from the 7 remaining players in the chess club in 7C3 ways.
Similarly, for the second school, 3 more players can be chosen for the contest from the 8 remaining players in the chess club in 8C3 ways.
Now, each team has 4 players for the contest; one of the schools has Rebecca and 3 more players, while the other one has Elise and 3 more players.
Now, Rebecca and Elise are already paired to play against each other. The remaining 3 pairs can be formed as follows:
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