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# help pls

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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
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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+0 Answers

#1
+2541
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