Difference between revisions of "2024 AMC 10A Problems/Problem 9"
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| − | + | ==Problem== | |
| + | In how many ways can <math>6</math> juniors and <math>6</math> seniors form <math>3</math> disjoint teams of <math>4</math> people so | ||
| + | that each team has <math>2</math> juniors and <math>2</math> seniors? | ||
| + | |||
| + | <math>\textbf{(A) }720\qquad\textbf{(B) }1350\qquad\textbf{(C) }2700\qquad\textbf{(D) }3280\qquad\textbf{(E) }8100</math> | ||
| + | |||
| + | == Solution 1== | ||
| + | The number of ways in which we can choose the juniors for the team are <math>{6\choose2}{4\choose2}{2\choose2}=15\cdot6\cdot1=90</math>. Similarly, the number of ways to choose the seniors are the same, so the total is <math>90\cdot90=8100</math>. But we must divide the number of permutations of three teams, since the order in which the teams were chosen never mattered, which is <math>3!</math>. Thus the answer is <math>\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}</math> | ||
| + | |||
| + | ~eevee9406 | ||
| + | ~small edits by NSAoPS | ||
| + | |||
| + | == Solution 2 == | ||
| + | Redacted because smn spammed. smn please rewrite this. ty | ||
| + | |||
| + | ~rcuber | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/l3VrUsZkv8I | ||
| + | |||
| + | == 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 == | ||
| + | |||
| + | https://youtu.be/AEd5tf1PJxk | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
| + | |||
| + | ==Video Solution by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=_o5zagJVe1U | ||
| + | |||
| + | ==Video Solution by Dr. David== | ||
| + | https://youtu.be/2EF0EDlxgkM | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2024|ab=A|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:04, 28 June 2025
Contents
[hide]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 because smn spammed. smn please rewrite this. ty
~rcuber
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. 
