combinations of four taken two at a time = 4C2 = 4!/[ 2! (4-2)! ]
= 4*3*2 /(2 * 2) = 6 games before the final
so 7
Four teams are playing in a tournament. Each team will play each other one more time before the top two teams play the championship game. How many games will be played?
Need some help with this mates.
8 answers
Whoops, you do not do it that way in grade 5
Make a grid
___A___B___C___D
A --- * --* --*
B----------*---*
C--------------*
D ------------- The stars are the games played in the first round
You see there are 6 of them
then the final is 7
Make a grid
___A___B___C___D
A --- * --* --*
B----------*---*
C--------------*
D ------------- The stars are the games played in the first round
You see there are 6 of them
then the final is 7
That did not line up right, but I hope you see the point
another way
A plays the other 3
B plays only C and D more because has already played A that is 2 more
C plays only D because has played A and B so one more
3+2+1 = 6 again
plus that last game = 7
A plays the other 3
B plays only C and D more because has already played A that is 2 more
C plays only D because has played A and B so one more
3+2+1 = 6 again
plus that last game = 7
Thanks mate. I owe ye big time.
(I'm not in 5th grade ;P)
(I'm not in 5th grade ;P)
You are welcome :)
At the volleyball championship tournament 7 teams are registered.If each teams must play each of the other teams how many games will be played math question.
In analysing the words of the question the words " one more time " to me imply that these teams played each other TWICE.
Therefore the number of games played will be : 3 + 2 + 1= 6games for the first round then another 6 games return matches plus the final game making a total of 13 games.
(i.e)
2(3+2+1)+ 1(Final game )
= 2(6)+1
= 12+1
= 13 games
Therefore the number of games played will be : 3 + 2 + 1= 6games for the first round then another 6 games return matches plus the final game making a total of 13 games.
(i.e)
2(3+2+1)+ 1(Final game )
= 2(6)+1
= 12+1
= 13 games