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Selection Game

The starting line-up for the Olympic basketball “Dream Team” is chosen from the following two groups:
Group A Group B
Johnson, Drexler, Bryant, Ewing Laettner, Robinson, James, Malone, Pippen

The following requirements must be meet:
Two players are chosen from Group A, and three from Group B.
James starts only if Bryant starts.
Drexler and Bryant do not both start.
If James starts, then Malone does not.
Exactly 3 of the four fast-break specialists—Johnson, Bryant, James, Pippen—must be chosen.
This problem is rather convoluted because not only are there direct conditions on the players, such as “Drexler and Bryant do not both start,” but there are also constraining numerical conditions, such as “exactly three fast-break specialists must be chosen.”

It is best to solve this problem without a diagram; however, we will still symbolize the conditions for clarity and easy reference. The condition “James starts only if Bryant starts” implies only that if James is starting then Bryant must be starting as well. So we symbolize it as James—>Bryant. The condition “Drexler and Bryant do not both start” means that if one starts then the other does not. So we symbolize it as Drexler—>~Bryant.* Students often misinterpret this condition to mean that neither of them starts. To state that neither starts, put both at the beginning of the sentence: Both Drexler and Bryant do not start.

The condition “if James starts, then Malone does not” is naturally symbolized as James—>~Malone. It tells us that if J starts then M does not, but tells us nothing when M does not start. Such a condition, where the two parts of an if-then statement do not similarly affect each other, is called a nonreciprocal condition. On the other hand, a condition such as James<—>~Malone affects J and M equally. In this case, we are told that if J starts then M does not as before, but we are told additionally that if M does not start then J does. It is important to keep the distinction between reciprocal and nonreciprocal relations clear; a common mistake is to interpret a nonreciprocal relation as reciprocal (see Unwarranted Assumptions, page 26). The remaining conditions cannot be easily written in symbol form, but we will paraphrase them in the schematic:
2 from Group A      3 from Group B
fast-break specialists: Johnson, Bryant, James, Pippen
3 fast-break specialists
Ewing, Laettner, Robinson are “wild”


Note: Ewing, Laettner, and Robinson are independent because there are no conditions that refer directly to them. We now turn to the questions.
If James starts, which of the following must also start?
  1. Malone or Johnson
  2. Drexler or Laettner
  3. Drexler or Johnson
  4. Johnson or Pippen
  5. Malone or Robinson
From the condition James—>Bryant, we know that if James starts, then Bryant must start as well.
Now both James and Bryant are fast-break specialists, and three of the four fast-break specialists must start.
So at least one of the remaining fast-break specialists—Johnson or Pippen—must also start.
The answer is (D).
All of the following pairs of players can start together EXCEPT:
  1. Ewing and Drexler
  2. James and Johnson
  3. Robinson and Johnson
  4. Johnson and Bryant
  5. Pippen and Malone
We shall use the method of indirect proof to solve this problem: That is, assume that a particular answer-choice is true. Then check whether it leads to a contradiction or an impossible situation. If so, it is the answer; if not, then select another answer-choice and repeat the process until a contradiction is found.
Begin with choice (A). Both Ewing and Drexler are from Group A, so the remaining 3 starters must be chosen from Group B.
Additionally, they must all be fast-break specialists since neither E nor D is—there are exactly 3 fast-break specialists.
But James and Pippen are the only fast-break specialists in Group B. So the third fast-break specialist cannot be chosen. The answer therefore is (A).
This type of question can be time consuming because you may have to check all the answer-choices—save these questions for last.
If the condition “Bryant starts only if Pippen doesn’t” is added to the other conditions, then which one of the following must be false?
  1. Johnson starts with Bryant
  2. Laettner starts with Malone
  3. Laettner starts with Bryant
  4. James starts with Robinson
  5. James starts with Bryant
This problem is both long and hard. Again, we use an indirect proof. Start with (A).
Both Johnson and Bryant are from Group A, and both are fast-break specialists.
So the remaining 3 starters must be chosen from Group B, one of which must be a fast-break specialist.
Now if James, Robinson, and Laettner are chosen, there will be three fast-break specialists and none of the initial conditions will be violated.
So (A) is not necessarily false; eliminate it. Next, we check (B).
Both Laettner and Malone are from Group B, and neither is a fast-break specialist.
So the three remaining starters must all be fast-break specialists, and two of them must be from Group A—Johnson and Bryant. This leaves only James and Pippen to choose from.
James cannot be chosen because Malone has already been chosen (James—>~Malone), and from the new condition Pippen cannot be chosen because Bryant has already been chosen.
Hence, the answer is (B).
If Malone starts, which one of the following is a complete and accurate list of the players from Group A any one of whom could also start?
  1. Johnson
  2. Johnson, Drexler
  3. Johnson, Bryant
  4. Johnson, Drexler, Bryant
  5. Johnson, Ewing, Bryant
James cannot start with Malone according to the condition James—>~Malone. To play three fast-break specialists, therefore, Johnson, Bryant, and Pippen are all required to start. Since both Johnson and Bryant are from Group A and exactly two players from that group start, these two players comprise the complete list of starters from Group A when Malone also starts. The answer is (C).
Which one of the following players must start?
  1. Pippen
  2. Johnson
  3. James
  4. Malone
  5. Bryant
Suppose Bryant does not start.
Then the 3 fast-break specialists must be Johnson, James, and Pippen.
But if James starts, then from the initial conditions Bryant must also start.
Hence, Bryant must always start.
The answer is (E).

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