Difference between revisions of "2013 AMC 10A Problems/Problem 24"
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Let us label the players of the first team <math>A</math>, <math>B</math>, and <math>C</math>, and those of the second team, <math>X</math>, <math>Y</math>, and <math>Z</math>. | Let us label the players of the first team <math>A</math>, <math>B</math>, and <math>C</math>, and those of the second team, <math>X</math>, <math>Y</math>, and <math>Z</math>. | ||
− | Let us first consider how to organize A's matches, AX, AX, AY, AY, <math>AZ</math>, and <math>AZ</math> | + | Let us first consider how to organize A's matches, <math>AX</math>, <math>AX</math>, <math>AY</math>, <math>AY</math>, <math>AZ</math>, and <math>AZ</math>. Because we have three duplicates, there are <math>\frac{6!}{2!*2!*2!} = 90</math> ways to organize A's matches. |
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+ | Now, consider <math>B</math> and <math>C</math>. WLOG assume that A's matches were <math>XXYYZZ</math>, as we will multiply by <math>90</math> the end anyways, and that, in the first round, <math>B</math> played <math>Y</math> and C played <math>Z</math>. | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=23|num-a=25}} | {{AMC10 box|year=2013|ab=A|num-b=23|num-a=25}} |
Revision as of 15:13, 8 February 2013
Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?
Solution
- Credit to the Math Jam for this solution
Let us label the players of the first team , , and , and those of the second team, , , and .
Let us first consider how to organize A's matches, , , , , , and . Because we have three duplicates, there are ways to organize A's matches.
Now, consider and . WLOG assume that A's matches were , as we will multiply by the end anyways, and that, in the first round, played and C played .
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AMC 10 Problems and Solutions |