Difference between revisions of "2024 AMC 10A Problems/Problem 9"
MRENTHUSIASM (talk | contribs) m (Deleted repetitive solution.) |
MRENTHUSIASM (talk | contribs) m |
||
| Line 1: | Line 1: | ||
==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 01:11, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
