Please help me solve this problem and explain how you did it!?

Question

There are seven basketball teams in Central Hoosier League. Ever year each team plays all the other teams in the league once. How many league games must be sceduled each year?

(Please answer in nth term or / probability (hint: ratio))

Answer 1

Each of the 7 teams play 6 other teams...but 7 x 6 = 42 would count every game twice, once for each team.

So it's 42/2 = 21 games.

Answer 2

Let's pick one team and call it A. A must play against all the other teams, so it will have to play 6 games.
Then lets take team B; it too must play all the other teams, for a total of 6, BUT we already assigned the match between team A and B to team A, so team B will account for 5 additional games that we have not already covered with team A.
Team C: 4 new games.
Team D: 3
Team E: 2
and Team F: 1 (against team G)
(Team G games are all listed already on the other teams agenda)

So the number of games will be 6+5+4+3+2+1 = 21

Answer 3

Consider the number of PAIRS of teams that can be chosen from these seven teams.

The first team in each pair can be chosen in 7 ways. That leaves 6 teams to choose from. So the second team in those pairs can be chosen from the remaining teams, that is in 6 ways. That makes 42 ordered pairs of type PQ, where for P and Q may be substituted ANY two different members of the full set: A, B, C, D, E, F, G. This includes the possibility that what we got in our arbitrarily ordered selection was the ordered pair BA as well as the ordered pair BA. (The letters A through G stand for the 7 different teams.)

So, in doing this, any given PAIR of teams could have appeared in the order (say) DE or in the order ED, that is, in those TWO different orders. But a PAIR is a PAIR, and each team only plays every other team ONCE.

So, choosing these possible pairings without regard to their order has actually DOUBLE-COUNTED: D and E appear both as DE and as ED. This is true of ALL the possible pairs.

So, to get the number of COMPLETELY INDEPENDENT PAIRS, we must divide the original 42 pairings (chosen without regard to the fact that a given pairing will appear TWICE) by exactly: 2.

So the number of DIFFERENT pairings, each individual pairing appearing only ONCE in the list, will be: 21.

Thus: 21 league games must be scheduled each year.

Live long and prosper.

Answer 4

1 2 3 4 5 6 7

Team 1 must play 2-7, which is 6 games.

Team 2 has already played Team 1, so they play 3-7, which is 5

Team 3 has already played 1 & 2, so they play 4-7, 4 games.

etc etc

I get 21 games total.