Difference between revisions of "2024 AMC 10A Problems/Problem 9"
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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? | 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= | + | ==Solution== |
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{1350}</math>. | 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{1350}</math>. | ||
~eevee9406 | ~eevee9406 | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2024|ab=A|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Revision as of 15:45, 8 November 2024
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
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, which is 
. Thus the answer is 
.
~eevee9406
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. 
