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` Six soccer teams are competing in a tournament in Waterloo. Every team is to play`

three games, each against a di erent team. (Note that not every pair of teams plays

a game together.) Judene is in charge of pairing up the teams to create a schedule

of games that will be played. Ignoring the order and times of the games, how many

di erent schedules are possible?

(A) 90 (B) 100 (C) 80 (D) 60 (E) 70

one year ago

Before we answer the given question, we determine the number of ways of choosing 3 objectsfrom 5 objects and the number of ways of choosing 2 objects from 5 objects.Consider 5 objects labelled B, C, D, E, F.The possible pairs are: BC, BD, BE, BF, CD, CE, CF, DE, DF, EF. There are 10 such pairs.The possible triples are: DEF, CEF, CDF, CDE, BEF, BDF, BDE, BCF, BCE, BCD. Thereare 10 such triples.(Can you see why there are the same number of pairs and triples?)Label the six teams A, B, C, D, E, F.We start by considering team A.Team A plays 3 games, so we must choose 3 of the remaining 5 teams for A to play. As we sawabove, there are 10 ways to do this.Without loss of generality, we pick one of these sets of 3 teams for A to play, say A plays B, Cand D.We keep track of everything by drawing diagrams, joining the teams that play each other witha line.Thus far, we haveThere are two possible cases now – either none of B, C and D play each other, or at least onepair of B, C, D plays each other.Case 1: None of the teams that play A play each otherIn the configuration above, each of B, C and D play two more games. They already play A andcannot play each other, so they must each play E and F.This givesNo further choices are possible.There are 10 possible schedules in this type of configuration. These 10 combinations come fromchoosing the 3 teams that play A.Case 2: Some of the teams that play A play each otherHere, at least one pair of the teams that play A play each other.Given the teams B, C and D playing A, there are 3 possible pairs (BC, BD, CD).We pick one of these pairs, say BC. (This gives 10 × 3 = 30 configurations so far.)It is now not possible for B or C to also play D. If it was the case that C, say, played D, thenwe would have the configurationAB C DE FHere, A and C have each played 3 games and B and D have each played 2 games. Teams Eand F are unaccounted for thus far. They cannot both play 3 games in this configuration asthe possible opponents for E are B, D and F, and the possible opponents for F are B, D andE, with the “B” and “D” possibilities only to be used once.A similar argument shows that B cannot play D.Thus, B or C cannot also play D. So we have the configurationHere, A has played 3 games, B and C have each played 2 games, and D has played 1 game.B and C must play 1 more game and cannot play D or A.They must play E and F in some order. There are 2 possible ways to assign these games (BEand CF, or BF and CE.) This gives 30 × 2 = 60 configurations so far.Suppose that B plays E and C plays F.So far, A, B and C each play 3 games and E, F and D each play 1 game.The only way to complete the configuration is to join D, E and F.Therefore, there are 60 possible schedules in this case.In total, there are 10 + 60 = 70 possible schedules.Option E

one year ago

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