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Difference between revisions of "2005 AMC 8 Problems/Problem 14"

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==Solution==
 
==Solution==
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Within each division, there are <math>_6 C _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.
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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.
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==Solution 2==
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Each team plays 10 games in its own division and 6 games against teams in the other division. So each of the 12 teams plays 16 conference games. Because each game involves two teams, there are <math>\frac{12\times 16}{2}=\boxed{96}</math> games scheduled. ~aopsav (Credit to AoPS Alcumus)
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=13|num-a=15}}
 
{{AMC8 box|year=2005|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 13:34, 2 May 2020

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?

$\textbf{(A)}\ 80\qquad\textbf{(B)}\ 96\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 192$

Solution

Within each division, there are $\binom {6}{2} = 15$ pairings, and each of these games happens twice. The same goes for the other division so that there are $4(15)=60$ games within their own divisions. The number of games between the two divisions is $(6)(6)=36$. Together there are $60+36=\boxed{\textbf{(B)}\ 96}$ conference games.

Solution 2

Each team plays 10 games in its own division and 6 games against teams in the other division. So each of the 12 teams plays 16 conference games. Because each game involves two teams, there are $\frac{12\times 16}{2}=\boxed{96}$ games scheduled. ~aopsav (Credit to AoPS Alcumus)

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

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