Difference between revisions of "2024 AMC 10A Problems/Problem 9"
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| + | =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 <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 | ||
Revision as of 15:44, 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
