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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 $\dfrac{3!}{4!}$???

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    $\begingroup$ The expression, though correct, is kind of silly. The probability R is chosen to be on her school's team is $\frac{4}{8}$. The corresponding probability for E is $\frac{4}{9}$. And given they are both chosen, the probability they play against each other is $\frac{1}{4}$. $\endgroup$ – André Nicolas Apr 30 '15 at 18:49
  • $\begingroup$ how come the probability they play against each other is 1/4 , not 1/4 * 1/4 ? $\endgroup$ – user235829 Apr 30 '15 at 18:50
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    $\begingroup$ Rebecca is equally likely to play against any of the $4$ members of the opposing team. $\endgroup$ – André Nicolas Apr 30 '15 at 18:51
  • $\begingroup$ Please see this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 30 '15 at 19:41
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The answer is:

$$\frac{\binom73\cdot\binom83\cdot3!}{\binom84\cdot\binom94\cdot4!}$$


A brief explanation:

  • The denominator counts the total number of combinations
  • The numerator counts the number of combinations in which the two sisters are paired
  • After we choose and pair the two sisters, we are left with one team of $7$ players from which we are to choose $3$ players, and one team of $8$ players from which we are to choose $3$ players
  • We then need to pair those two groups of $3$ players each, as the two sisters are already paired

The total number of combinations:

  • The number of ways to choose $4$ out of $8$ players from the first team is $\binom84$
  • The number of ways to choose $4$ out of $9$ players from the second team is $\binom94$
  • The number of ways to order one of those groups of $4$ players in any possible manner is $4!$

By ordering one group in any possible manner, we are essentially counting the number of ways to pair the players of that group with the players of the other group.


The number of combinations in which the two sisters are paired:

  • The number of ways to choose $3$ out of $7$ players from the first team is $\binom73$
  • The number of ways to choose $3$ out of $8$ players from the second team is $\binom83$
  • The number of ways to order one of those groups of $3$ players in any possible manner is $3!$

By ordering one group in any possible manner, we are essentially counting the number of ways to pair the players of that group with the players of the other group.

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  • $\begingroup$ A good explanation how to derive the book formula. As others have shown, there is a much easier route to the result, but that was not the question. $\endgroup$ – Ross Millikan Apr 30 '15 at 19:42
  • $\begingroup$ @RossMillikan: Yep, I followed OP's question in a straightforward manner. $\endgroup$ – barak manos Apr 30 '15 at 19:46
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Without loss of generality, let's assume that Rebecca is in the school that has 8 on their team. (i.e. If not then change the names).

The Probability of Rebecca being chosen is 1/2.

The probability that the selected opponent will be her sister is 1/9.

You have to walk the last step on your own.

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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 $(7C3×8C3×3!)/(8C4×9C4×4!)$.

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    $\begingroup$ Please, use Mathjax! $\endgroup$ – Aqua May 12 at 20:57

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