Difference between revisions of "2005 AMC 8 Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | Within each division, there are <math>\binom {6}{2} = 15</math> pairings, and each of these games happens twice. The same goes for the other division so that there are <math>4(15)=60</math> games within their own divisions. The number games between the two divisions is <math>(6)(6)=36</math>. Together there are <math>60+36=\boxed{\textbf{(B)}\ 96}</math> conference games. | + | Within each division, there are <math>\binom {6}{2} = 15</math> pairings, and each of these games happens twice. The same goes for the other division so that there are <math>4(15)=60</math> games within their own divisions. The number of games between the two divisions is <math>(6)(6)=36</math>. Together there are <math>60+36=\boxed{\textbf{(B)}\ 96}</math> conference games. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=13|num-a=15}} | {{AMC8 box|year=2005|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:18, 28 October 2019
Problem
The Little Twelve Basketball Conference has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many conference games are scheduled?
Solution
Within each division, there are 
pairings, and each of these games happens twice. The same goes for the other division so that there are 
games within their own divisions. The number of games between the two divisions is 
. Together there are 
conference games.
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
