Difference between revisions of "2024 AMC 10A Problems/Problem 9"
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== Solution 1== | == Solution 1== | ||
| − | The number of ways in which we can choose the juniors for the team are <math>{6\choose2}{4\choose2}{2\choose2}=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, which is <math>3!</math>. Thus the answer is <math>\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}</math>. | + | 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 | ~eevee9406 | ||
| + | ~small edits by NSAoPS | ||
| + | |||
| + | == Solution 2 == | ||
| + | Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is <math>{6\choose2}{6\choose2} = 15\cdot 15 = 225</math>. Now, only four juniors and four seniors are left to choose the second team. Thus, the second team can be formed in <math>{4\choose2}{4\choose2} = 6\cdot 6 = 36</math> ways. There are now only two juniors and two seniors left, so the third team can only be formed in one way. Thus, the total number of ways in which we can choose two juniors and two teams for three distinguishable teams is <math>225\cdot 36 = 8100</math> ways. However, the problem does not require the teams to be distinguishable. Therefore, we must divide by <math>3!=6</math> to find the answer for three indistinguishable teams. | ||
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| + | The answer is <math>\frac{8100}{6} =\boxed{\textbf{(B) }1350}</math>. | ||
| + | |||
| + | ~jjjxia | ||
== Video Solution by Pi Academy == | == Video Solution by Pi Academy == | ||
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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 == | ||
| + | |||
| + | https://youtu.be/AEd5tf1PJxk | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
| + | |||
| + | ==Video Solution by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=_o5zagJVe1U | ||
==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}} | ||
Latest revision as of 19:48, 17 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
Solution 2
Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is 
. Now, only four juniors and four seniors are left to choose the second team. Thus, the second team can be formed in 
ways. There are now only two juniors and two seniors left, so the third team can only be formed in one way. Thus, the total number of ways in which we can choose two juniors and two teams for three distinguishable teams is 
ways. However, the problem does not require the teams to be distinguishable. Therefore, we must divide by 
to find the answer for three indistinguishable teams.
The answer is 
.
~jjjxia
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
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. 
