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AMC 10

 
 
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Re: AMC 10
ReynosoDavid - 2020年12月15日 Tuesday 11:46
 

We can take a look at this problem in detail during the live session today. Please remember to let us know what you have tried so far so we can help you better.

In the meantime, here's a hint to get you started.

Say the players from Central High School are $A_1$, $A_2$, and $A_3$, and the players from Northern High School are $B_1$, $B_2$, and $B_3$. Let the players from CH choose their opponents. If $A_1$ chooses first they have $3$ players to choose from, then $A_2$ has $2$ players to choose from, and $A_3$ plays against the remaining player. Thus, there are $3! = 6$ ways to choose each round. Note this is the same as the number of rounds, and each player plays each player from the opposing team twice. Note the order of the rounds matters. Now, is this the only way to generate the $6$ rounds?