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 <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== | == 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{\textbf{(B) }1350}</math>. |
~eevee9406 | ~eevee9406 |
Revision as of 02:11, 9 November 2024
Problem
In how many ways can juniors and
seniors form
disjoint teams of
people so
that each team has
juniors and
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
Video Solution 1 by Power Solve
https://youtu.be/j-37jvqzhrg?si=IBSPzNSvdIodGvZ7&t=1145
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.