# Six soccer teams are competing in a tournament in Waterloo. Every team is to playthree games, each against a di erent team. (Note that not every pair of teams playsa game together.) Judene is in charge of pairing up the teams to create a scheduleof games that will be played. Ignoring the order and times of the games, how manydi erent schedules are possible?(A) 90 (B) 100 (C) 80 (D) 60 (E) 70

Arun
25750 Points
6 years ago
Before we answer the given question, we determine the number of ways of choosing 3 objects
from 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. There
are 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 saw
above, 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, C
and D.
We keep track of everything by drawing diagrams, joining the teams that play each other with
a line.
Thus far, we have

There are two possible cases now – either none of B, C and D play each other, or at least one
pair of B, C, D plays each other.
Case 1: None of the teams that play A play each other
In the configuration above, each of B, C and D play two more games. They already play A and
cannot play each other, so they must each play E and F.
This gives
No further choices are possible.
There are 10 possible schedules in this type of configuration. These 10 combinations come from
choosing the 3 teams that play A.
Case 2: Some of the teams that play A play each other
Here, 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, then
we would have the configuration
A
B C D
E F
Here, A and C have each played 3 games and B and D have each played 2 games. Teams E
and F are unaccounted for thus far. They cannot both play 3 games in this configuration as
the possible opponents for E are B, D and F, and the possible opponents for F are B, D and
E, 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 configuration

Here, 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 (BE
and 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