- 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?

Guest Aug 8, 2022

#1**0 **

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

#1**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