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

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== Solution 2 ==
 
== Solution 2 ==
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Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is <math>{6\choose2}{6\choose2} = 15\cdot 15 = 225</math>. Now, only four juniors and four seniors are left to choose the second team. Thus, the second team can be formed in <math>{4\choose2}{4\choose2} = 6\cdot 6 = 36</math> ways. There are now only two juniors and two seniors left, so the third team can only be formed in one way. Thus, the total number of ways in which we can choose two juniors and two teams for three distinguishable teams is <math>225\cdot 36 = 8100</math> ways. However, the problem does not require the teams to be distinguishable. Therefore, we must divide by <math>3!=6</math> to find the answer for three indistinguishable teams.
+
Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is <math>{6\choose2}{6\choose2} = 15\cdot 15 = 225</math>. Now, only four juniors and four seniors are left to choose the second team. Thus, the second team can be formed in <math>{4\choose2}{4\choose2} = 6\cdot 6 = 36</math> ways. There are now only two juniors and two seniors left, so the third team can only be formed in one way. Thus, the total number of ways in which we can choose two juniors and two seniors for three distinguishable teams is <math>225\cdot 36 = 8100</math> ways. However, the problem does not require the teams to be distinguishable. Therefore, we must divide by <math>3!=6</math> to find the answer for three indistinguishable teams.
  
 
The answer is <math>\frac{8100}{6} =\boxed{\textbf{(B) }1350}</math>
 
The answer is <math>\frac{8100}{6} =\boxed{\textbf{(B) }1350}</math>
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~edit by yanes04 (wrong number)
 
~edit by yanes04 (wrong number)
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~ edit by RoyalPawn38
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==Video Solution==
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https://youtu.be/l3VrUsZkv8I
  
 
== Video Solution by Pi Academy ==
 
== Video Solution by Pi Academy ==
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https://www.youtube.com/watch?v=_o5zagJVe1U
 
https://www.youtube.com/watch?v=_o5zagJVe1U
  
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==See also==
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==Video Solution by Dr. David==
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{{AMC10 box|year=2024|ab=A|num-b=8|num-a=10}}
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https://youtu.be/2EF0EDlxgkM
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== Video solution by TheNeuralMathAcademy ==
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https://www.youtube.com/watch?v=4b_YLnyegtw&t=1430s
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==See Also==
 +
{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}}
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* [[AMC 10]]
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* [[AMC 10 Problems and Solutions]]
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* [[Mathematics competitions]]
 +
* [[Mathematics competition resources]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:44, 19 August 2025

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?

$\textbf{(A) }720\qquad\textbf{(B) }1350\qquad\textbf{(C) }2700\qquad\textbf{(D) }3280\qquad\textbf{(E) }8100$

Solution 1

The number of ways in which we can choose the juniors for the team are ${6\choose2}{4\choose2}{2\choose2}=15\cdot6\cdot1=90$. 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, since the order in which the teams were chosen never mattered, which is $3!$. Thus the answer is $\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}$

~eevee9406 ~small edits by NSAoPS

Solution 2

Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is ${6\choose2}{6\choose2} = 15\cdot 15 = 225$. Now, only four juniors and four seniors are left to choose the second team. Thus, the second team can be formed in ${4\choose2}{4\choose2} = 6\cdot 6 = 36$ ways. There are now only two juniors and two seniors left, so the third team can only be formed in one way. Thus, the total number of ways in which we can choose two juniors and two seniors for three distinguishable teams is $225\cdot 36 = 8100$ ways. However, the problem does not require the teams to be distinguishable. Therefore, we must divide by $3!=6$ to find the answer for three indistinguishable teams.

The answer is $\frac{8100}{6} =\boxed{\textbf{(B) }1350}$

~jjjxi

~edit by yanes04 (wrong number)

~ edit by RoyalPawn38

Video Solution

https://youtu.be/l3VrUsZkv8I

Video Solution by Pi Academy

https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv

Video Solution 1 by Power Solve

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

Video Solution by Daily Dose of Math

https://youtu.be/AEd5tf1PJxk

~Thesmartgreekmathdude

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=_o5zagJVe1U

Video Solution by Dr. David

https://youtu.be/2EF0EDlxgkM

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=1430s

See Also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
2023 AMC 10B Problems 		Followed by
2024 AMC 10B Problems
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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