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?

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 $\frac{8100}{3!}=\frac{8100}{6}=\boxed{1350}$ .

~eevee9406

Solution 2(Pretty Much The Exact Same Solution As #1)

Let the three teams be team A, team B, and team C. There are $\binom{6}{2} = 15$ ways to choose the two seniors for team A. Then, out of the remaining four, there are $\binom{4}{2} = 6$ ways to choose two seniors for team B. Then, the remaining two seniors must be on team C. Thus, there are $15 \cdot 6 = 90$ ways to choose seniors for all the teams.

Similarly, there are $90$ ways to choose juniors for the three teams, so there are $90 \cdot 90 = 8100$ ways in total. However, since the teams are symmetric, we must divide by $3! = 6$ , so the final answer is $\frac{8100}{6} = \boxed{\textbf{(B)} 1350}$

~andliu766

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2024 AMC 10A Problems/Problem 9

Contents

Problem

Solution 1

Solution 2(Pretty Much The Exact Same Solution As #1)

Video Solution 1 by Power Solve

See also