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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 21"
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There are <math>{8\choose 2}-12=16</math> games.
 
There are <math>{8\choose 2}-12=16</math> games.
  
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== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 20|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 22|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
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[[Category:Introductory Combinatorics Problems]]

Revision as of 10:51, 23 July 2006

Problem

Suppose that each pair of eight tennis players either played exactly one game last week or did not play at all. Each player participated in all but 12 games. How many games were played among the eight players?

$\mathrm{(A) \ }10 \qquad \mathrm{(B) \ }12 \qquad \mathrm{(C) \ }14 \qquad \mathrm{(D) \ }16 \qquad \mathrm{(E) \ }18$

Solution

There are ${8\choose 2}-12=16$ games.

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