That is in a league where every team plays twice against the same opponent (Home and away games) So with only 2 teams it is only 2 games and with 3 teams 6 games. How much for n teams?
2006-07-10 07:21:35 · 10 answers · asked by Jan 3 in Science & Mathematics ➔ Mathematics
Assume that there are 4 teams: a,b,c, and d and they play each other once (multiply final answer by 2 to get they play each team twice).
Therefore:
a plays: b, c, and d
b plays: c and d (they already played a)
c plays: d (they already played a and b)
d plays: nobody (they already played a, b, and c)
Therefore the number of games that were played are 1+2+3=6
In general, if there are n teams, then the number of games played would be 1+2+3+4+5+ . . +(n-1). This sum is known to equal (n-1)((n-1)+1)/2=n(n-1)/2.
So, if there are n teams, and they play each team once, there will be n(n-1)/2 games.
If they play each team m times, there will be m•n(n-1)/2.
In your case, they play each team twice (so m=2) and the number of games played is 2n(n-1)/2=n(n-1).
2006-07-10 08:23:39 · answer #1 · answered by Eulercrosser 4 · 6⤊ 0⤋
Every game played is a home game for exactly one of the teams involved. Count the home games and you have the answer.
Each of the n teams plays the other (n-1) teams at home, so they play n-1 home games. There are n teams so there are n(n-1) home games.
Another way to see it is to use combinations. How may pairings of two teams are there? nC2 = n!/2!(n-2) = n(n-1)/2. Each pairing plays two games so there are n(n-1)games played.
2006-07-10 08:25:06 · answer #2 · answered by rt11guru 6 · 0⤊ 0⤋
If eact team play each other once then the formula is n(n-1)/2
If each team play each other twice then the formula is n(n-1)
Example
There are 20 teams in the English premier league and they play each other twice, once at home and once away, so there are n(n-1) total games, which is 20 x 19 = 380
2006-07-10 23:46:00 · answer #3 · answered by Anonymous · 0⤊ 0⤋
The number of games that one team plays is N-1.
(They play everybody but themselves). ----- So N teams play N(N-1) games, if they play one Game against each other team.
For more games, just multiply N(N-1) times the number of games (call it G) that you play against each team.
For two games against each team it would be 2 X N(N-1)
For any number of games: The league plays G X N(N-1)
2006-07-10 07:50:20 · answer #4 · answered by Anonymous · 0⤊ 0⤋
2 teams = 2 games
3 teams = 6 games
4 teams = 12 games
5 teams = 20 games
6 teams = 30 games
7 teams = 42 games
8 teams = 56 games
As you add 1 team it goes up by 4,6,8,10,12,14,16 etc.
This is an example of (n sqrd - n). N being the number of teams.
2006-07-10 07:35:28 · answer #5 · answered by Jeff U 4 · 0⤊ 0⤋
The formula n(n-1) is correct for the total number of games played home and away, not just the home games as someone said. This is because one persons home game is the other sides away game so it is already included.
2006-07-10 09:21:01 · answer #6 · answered by brainyandy 6 · 0⤊ 0⤋
n(n-1) or n^2 - n total games would be on the schedule
2006-07-10 09:53:00 · answer #7 · answered by jimbob 6 · 0⤊ 0⤋
Double it and subtract 2.
2006-07-10 07:36:46 · answer #8 · answered by Roxy 6 · 0⤊ 0⤋
i no!
3xy+087 (bearing of the philosophical existence) / 6uclfhj;xc;g dkh=0u
u
hj
mj
6rp
fg
lnm[ytl# 'fkj'yt,kyf
= 1
duh i cant believe u cudnt work that out!
2006-07-10 07:25:54 · answer #9 · answered by ChipZ 2 · 0⤊ 0⤋
n*n - n
think of it as a grid n x n minus thi diagonal.
2006-07-10 07:26:46 · answer #10 · answered by gjmb1960 7 · 0⤊ 0⤋