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

Revision as of 09:53, 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?

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

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

~eevee9406

Solution 3

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
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 ~eevee9406
+== Solution 3 ==
 == Video Solution 1 by Power Solve ==

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