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There are $5$ girls and $5$ boys in a chess club. The club holds a round-robin tournament in which every player plays against every other player exactly once.

In how many games does a girl play against another girl?

User Skfp
by
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2 Answers

7 votes

Answer:

10

Explanation:

There are $5$ girls and $5$ boys in a chess club. The club holds a round-robin tournament-example-1
User Adam Amin
by
5.1k points
3 votes

Let's call the five girls
g_1, g_2, g_3, g_4 \text {and } g_5.


Since every player plays against every other player exactly once,
g_1 has to play against
g_2, g_3, g_4 \text {and } g_5. This are four maches.



g_2 has already played against
g_1, so she has to play against
g_3, g_4 \text {and } g_5. This are three matches.



g_3 has already played against
g_1 \text{ and } g_2, so she has to play against
g_4 \text {and } g_5. This are two matches.



g_4 has already played against
g_1, g_2 \text{ and } g_3, so she has to play against
g_5. This is one match.


So, the total is
4+3+2+1 = 10 matches.

User Onof
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5.7k points