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Difference between revisions of "2024 AMC 10A Problems/Problem 9"

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==Problem==
 
==Problem==
 
In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so that each team has 2 juniors and 2 seniors?
 
In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so that each team has 2 juniors and 2 seniors?
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== Solution 1 ==
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== Solution==
 
The number of ways in which we can choose the juniors for the team are <math>{6\choose2}{4\choose2}{2\choose2}=90</math>. Similarly, the number of ways to choose the seniors are the same, so the total is <math>90\cdot90=8100</math>. But we must divide the number of permutations of three teams, which is <math>3!</math>. Thus the answer is <math>\frac{8100}{3!}=\frac{8100}{6}=\boxed{1350}</math>.
 
The number of ways in which we can choose the juniors for the team are <math>{6\choose2}{4\choose2}{2\choose2}=90</math>. Similarly, the number of ways to choose the seniors are the same, so the total is <math>90\cdot90=8100</math>. But we must divide the number of permutations of three teams, which is <math>3!</math>. Thus the answer is <math>\frac{8100}{3!}=\frac{8100}{6}=\boxed{1350}</math>.
  
 
~eevee9406
 
~eevee9406
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== Solution 2(Pretty Much The Exact Same Solution As #1) ==
 
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Let the three teams be team A, team B, and team C. There are <math>\binom{6}{2} = 15</math> ways to choose the two seniors for team A. Then, out of the remaining four, there are <math>\binom{4}{2} = 6</math> ways to choose two seniors for team B. Then, the remaining two seniors must be on team C. Thus, there are <math>15 \cdot 6 = 90</math> ways to choose seniors for all the teams.
 
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Similarly, there are <math>90</math> ways to choose juniors for the three teams, so there are <math>90 \cdot 90 = 8100</math> ways in total. However, since the teams are symmetric, we must divide by <math>3! = 6</math>, so the final answer is <math>\frac{8100}{6} = \boxed{\textbf{(B)} 1350}</math>
 
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~andliu766
 
  
 
== Video Solution 1 by Power Solve ==
 
== Video Solution 1 by Power Solve ==

Revision as of 01:10, 9 November 2024

Problem

In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so that each team has 2 juniors and 2 seniors?

Solution

The number of ways in which we can choose the juniors for the team are ${6\choose2}{4\choose2}{2\choose2}=90$. Similarly, the number of ways to choose the seniors are the same, so the total is $90\cdot90=8100$. But we must divide the number of permutations of three teams, which is $3!$. Thus the answer is $\frac{8100}{3!}=\frac{8100}{6}=\boxed{1350}$.

~eevee9406

Video Solution 1 by Power Solve

https://youtu.be/j-37jvqzhrg?si=IBSPzNSvdIodGvZ7&t=1145

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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