2024 AMC 10A Problems/Problem 9

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 ${6\choose2}{4\choose2}{2\choose2}=90$ . 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 $\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}$ .

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Video Solution 1 by Power Solve

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

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2024 AMC 10A Problems/Problem 9

Contents

Problem

Solution 1

Video Solution 1 by Power Solve

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