Difference between revisions of "2024 AMC 10A Problems/Problem 9"
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==Problem== | ==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? | 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 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{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 | ||
| + | |||
| + | == Solution 2 == | ||
| + | Let the three teams be team A, team B, and team C. There are <math>\binom{6}{2} = 15</math> ways to choose the two seniors for team A. Then, out of the remaining four, there are <math>\binom{4}{2} = 6</math> ways to choose two seniors for team B. Then, the remaining two seniors must be on team C. Thus, there are <math>15 \cdot 6 = 90</math> ways to choose seniors for all the teams. | ||
| + | |||
| + | Similarly, there are <math>90</math> ways to choose juniors for the three teams, so there are <math>90 \cdot 90 = 8100</math> ways in total. However, since the teams are symmetric, we must divide by <math>3! = 6</math>, so the final answer is <math>\frac{8100}{6} = \boxed{\textbf{(B)} 1350}</math> | ||
| + | |||
| + | ~andliu766 | ||
| + | |||
==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 16:22, 8 November 2024
Contents
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 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, which is 
. Thus the answer is 
.
~eevee9406
Solution 2
Let the three teams be team A, team B, and team C. There are 
ways to choose the two seniors for team A. Then, out of the remaining four, there are 
ways to choose two seniors for team B. Then, the remaining two seniors must be on team C. Thus, there are 
ways to choose seniors for all the teams.
Similarly, there are 
ways to choose juniors for the three teams, so there are 
ways in total. However, since the teams are symmetric, we must divide by 
, so the final answer is 
~andliu766
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. 
