In a twelve team bowling league, if every team plays every other team four times during the season, how many games must be scheduled?
2007-01-12 14:17:28 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Each team must play 44 games (4 games against each of the other 11 teams). Thus, the total number of games which must be scheduled is 44 for the first team, 40 for the second (b/c the four with the first team have already been scheduled), 36 for the third, etc.
So, 44 + 40 + 36 + ... + 4 = 44*6 = 264
2007-01-12 14:23:11 · answer #1 · answered by JasonM 7 · 3⤊ 0⤋
Think of a single team. They will play 11 other teams, 4 times. That's a total of 44 games per team. However, this double counts the number of games because if team A plays B, that is the same as team B plays A.
So take 12 x 22 = 264
Another way to look at this is to figure the number of ways to pick two teams to play. This is 12 "choose" 2 which equates to 12 x 11 / 2 = 66 ways to match teams. But you said they meet 4 times during the season. 4 x 66 = 264.
Either way you get the same answer.
2007-01-12 22:24:34 · answer #2 · answered by Puzzling 7 · 0⤊ 0⤋
In your league, each team plays 44 games. There are 12 team in the league, so simply multiply 44 and 6. The reason it is 6 is because it takes two team to play one game. Multiplying 44 and 6 produces 264 games that have to be scheduled in order for your league to work.
2007-01-13 09:55:10 · answer #3 · answered by CSUFGrad2006 5 · 0⤊ 0⤋
Every team plays 11 other teams 4 times, or plays 44 "games".
12 * 44 is 528 games, but each game has two teams, so the answer is 528/2 or 264.
2007-01-12 22:23:25 · answer #4 · answered by teacher2006 3 · 0⤊ 0⤋
isnt this supposed to be homework or something?
2007-01-12 22:27:08 · answer #5 · answered by . 4 · 0⤊ 1⤋