6. There are 6 teams in a school football league. If each team plays each other team once during the season, how many games are played during the season? Explain your answer.
| 0 | Answer does not address task, or is unrelated or inappropriate. Answer contains nothing correct. |
| 1 | Answer is partially correct. Answer contains something correct in the process of calculating the number of games played or, a correct answer of 15 games is given without explanation or with unclear explanation. |
| 2 | Answer is nearly correct. The answer or explanation contains a minor error, omission, or lack of clarity. |
| 3 | Answer and explanation are complete and correct. |
6. There are 6 teams in a school football league. If each team plays each other team once during the season, how many games are played during the season? Explain your answer.
If there are 6 teams and they play each other once Then there are twelve football games played.
0 --Nothing correct.
30 games
6 teams have to play 5 other teams because they can't play themselves.
1 -- The answer of 30 is incorrect, but there is a correct beginning strategy (multiplying 6 teams by 5 teams).
15 games
each team gets a turn playing each other. The first team gets to play them first, then the second, third, and so on.
1 -- A correct answer is given with an explanation which shows a major lack of clarity.
15 games
There are 6 teams.
Teams 1, 2, 3, 4, 5, 6
| 1 & 2 = 1 | 2 & 3 = 1 | 3 & 4 = 1 |
| 1 & 3 = 1 | 2 & 4 = 1 | 3 & 5 = 1 |
| 1 & 4 = 1 | 2 & 5 = 1 | 3 & 6 = 1 |
| 1 & 5 = 1 | 4 & 5 = 1 | 5 & 6 = 1 |
| 4 & 6 = 1 |
2 -- A correct answer is given. Explanation lacks verbal detail/full clarity.
16 games
| a | ab | ac | ad | ae | af |
| b | bc | bd | be | bf | |
| c | cd | ce | ce | cf | |
| d | de | df | |||
| e | ef | ||||
| f |
I got this answer by counting how many combinations of games could be played.
2 -- A minor error is made: the combination ce is counted twice.
15
If one team plays all the others that would be 5 games played. Then the 2nd team has to play all of the other teams except team 1 which they have already played. That makes 4 games played. The team 3 would have already played 2 of the other teams so that leaves 3 more games & so on. Then you add them all up. 5 + 4 + 3 + 2 + 1 = 15.
3 -- Answer and explanation are complete and correct.