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  • A sports conference is split into two divisions, Division A and Division B. During a season, each team in Division A plays every other team in Division A three times, each team in Division B plays every other team in Division B twice, and each team in Division A plays each team in Division B once.

    If there are a total of 23  teams in the conference and 454  games are played during the season, how many teams are there in each division?
 Aug 8, 2022

Best Answer 

 #1
avatar+2446 
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Here we go again... my computer froze, so now I need to retype everything

 

Let a and b represent the number of students in divisions A and B, respectively. 

 

There are \(3 \times {a \choose 2} = {3a(a-1) \over 2}\) ways to pair the teams in Division A, and \(2 \times {b \choose 2} = b(b-1)\) ways to pair the teams in Division B.

 

There are also \(ab \) ways to pair the students from different divisions. 

 

This means we can form the equation \({3a(a-1) \over 2} + b(b-1) + ab = 454 \ \ \ \ (i)\)  

 

We also know that \(a + b = 23 \ \ \ \ (ii)\)

 

Solving the system and choosing the integer solutions gives us, in terms of \((a,b)\)\(\color{brown}\boxed{(13, 10)}\)

 Aug 8, 2022
 #1
avatar+2446 
0
Best Answer

Here we go again... my computer froze, so now I need to retype everything

 

Let a and b represent the number of students in divisions A and B, respectively. 

 

There are \(3 \times {a \choose 2} = {3a(a-1) \over 2}\) ways to pair the teams in Division A, and \(2 \times {b \choose 2} = b(b-1)\) ways to pair the teams in Division B.

 

There are also \(ab \) ways to pair the students from different divisions. 

 

This means we can form the equation \({3a(a-1) \over 2} + b(b-1) + ab = 454 \ \ \ \ (i)\)  

 

We also know that \(a + b = 23 \ \ \ \ (ii)\)

 

Solving the system and choosing the integer solutions gives us, in terms of \((a,b)\)\(\color{brown}\boxed{(13, 10)}\)

BuilderBoi Aug 8, 2022

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