Difference between revisions of "2024 AMC 10A Problems/Problem 9"
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https://youtu.be/j-37jvqzhrg?si=IBSPzNSvdIodGvZ7&t=1145 | https://youtu.be/j-37jvqzhrg?si=IBSPzNSvdIodGvZ7&t=1145 | ||
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| + | == Video Solution by Daily Dose of Math == | ||
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| + | https://youtu.be/AEd5tf1PJxk | ||
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| + | ~Thesmartgreekmathdude | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=8|num-a=10}} | {{AMC10 box|year=2024|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:36, 9 November 2024
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
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
