That depends on how you organize your data:
Firstly, there are only two possible scores, as the game currently stands - commencing 2011: a goal = 6 points; and a 'behind' = one point.
So for any two games:
Each game will have 2 possible opening scores. So the total possible outcomes might be:
Game A: 1 goal and Game B: 1 goal
Game A: 1 goal and Game B: 1 point
Game A: 1 point and Game B: 1 goal
Game A: 1 point and Game B: 1 point
If you are talking about Gaelic Football and Australian rules teams, then the answer is no. The first time Gaelic Football and Australian rules teams played each other was in 1967. A Meath team played an Australian team.
never England rule
The Melbourne Football Club which was founded on 14 May 1859.
IF YOU HAVE ONE GAME, YOU HAVE A POSIBILITY OF 3 OUTCOMES, IF YOU HAVE 2 GAMES IT GO'S UP TO 9 POSIBILE OUTCOMES. FIRST ANSWER 1X3=3 SECOND IS 3X3=9 FOR 10 GAMES IT IS 3X3 TEN TIMES 3X3X3X3X3X3X3X3X3X3=59,049 POSSIBLE COMBINATIONS (I THINK)HA HA
1859 between Melbourne Grammar school and Scotch College.
Brisbane And Port adielade were the first to teams Brisbane And Port adielade were the first to teams
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yes they are...the English football assocation wrote the rules for football in 1863
From 2001-2010 it was played at the end of September, but in 2011 it will be played on the first week of October.
Although there is a punter named Ben Graham (who started out in Australian football), you're probably looking for Otto Graham, who is in the Football Hall of Fame.
The first match was played in 1859 between Melbourne Grammar school and Scotch College.
To calculate the total number of possible outcomes in 4 different football games with 8 total teams, you would multiply the number of teams in each game together. In this case, there are 8 teams in the first game, 7 teams in the second game (as one team has already been used), 6 teams in the third game, and 5 teams in the fourth game. Therefore, the total number of possible outcomes would be 8 x 7 x 6 x 5 = 1,680 possible outcomes.