+2666 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 \((a,b)\), \(\color{brown}\boxed{(13, 10)}\)
+2666 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 \((a,b)\), \(\color{brown}\boxed{(13, 10)}\)
