Difference between revisions of "2024 AMC 10A Problems/Problem 9"
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Revision as of 21:03, 28 June 2025
Contents
Problem
In how many ways can 
juniors and seniors form 
disjoint teams of 
people so
that each team has 
juniors and seniors?
Solution 1
The number of ways in which we can choose the juniors for the team are 
. Similarly, the number of ways to choose the seniors are the same, so the total is 
. But we must divide the number of permutations of three teams, since the order in which the teams were chosen never mattered, which is 
. Thus the answer is 
~eevee9406 ~small edits by NSAoPS
Solution 2
Redacted by rcuber because smn spammed. smn please rewrite this. ty
Video Solution
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
~Thesmartgreekmathdude
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=_o5zagJVe1U
Video Solution by Dr. David
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
