How many games are played in a women's soccer conference if there are eight teams and all teams play each other twice?
8 (teams) x 7 (possible opponents of each team) / 2 = 28
The division by two is necessary to avoid counting each game twice.
I should have multiplied by 2 since each pair of teams plays twice. That makes the final number 56
To find the total number of games played in a women's soccer conference with eight teams playing each other twice, you need to use a combination of permutations and simple arithmetic.
Step 1: Determine the number of unique matchups between two teams. In a conference with eight teams, the first team can play against the remaining seven teams. The second team can then play against six remaining teams, and so on. To calculate this, you can use the formula for permutations without repetition: P(n, r) = n! / (n - r)!
In this case, P(8, 2) = 8! / (8 - 2)! = 8! / 6! = 8 * 7 = 56.
Step 2: Each unique matchup happens twice because every team plays against each other twice. So, multiply the number of unique matchups by 2.
56 * 2 = 112.
Therefore, in a women's soccer conference with eight teams playing against each other twice, a total of 112 games will be played.