Difference between revisions of "2024 AMC 10A Problems/Problem 9"

Revision as of 18:08, 26 June 2025

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?

$\textbf{(A) }720\qquad\textbf{(B) }1350\qquad\textbf{(C) }2700\qquad\textbf{(D) }3280\qquad\textbf{(E) }8100$

Solution 1

The number of ways in which we can choose the juniors for the team are ${6\choose2}{4\choose2}{2\choose2}=15\cdot6\cdot1=90$ . Similarly, the number of ways to choose the seniors are the same, so the total is $90\cdot90=8100$ . But we must divide the number of permutations of three teams, since the order in which the teams were chosen never mattered, which is $3!$ . Thus the answer is $\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}$

~eevee9406 ~small edits by NSAoPS

Solution 2

Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is chooREREORIWEPOIRCMPSFKSDMFLCKSJFOEPSKLDMFJOFEWM4 ways. There are now only two juniors and two seniors left, so the thirM9 FID9U90WU JT9 94 TU092U9T4U2092409294UTEJD OXFDNCVJ X the answer for 204290348920483209490 indistinguishable teams.

The answer is IMMA MONKEY OOO AAA EEEE HEHEHEHEHAHAHAHAH :D

Video Solution

https://youtu.be/l3VrUsZkv8I

Video Solution by Pi Academy

https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv

Video Solution 1 by Power Solve

https://youtu.be/j-37jvqzhrg?si=IBSPzNSvdIodGvZ7&t=1145

Video Solution by Daily Dose of Math

https://youtu.be/AEd5tf1PJxk

~Thesmartgreekmathdude

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=_o5zagJVe1U

Video Solution by Dr. David

https://youtu.be/2EF0EDlxgkM

@@ Line 12: / Line 12: @@
 == Solution 2 ==
-Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is <math>{6\choose2}{6\choose2} = 15\cdot 15 = 225</math>. Now, only four juniors and four seniors are left to choose the second team. Thus, the second team can be formed in <math>{4\choose2}{4\choose2} = 6\cdot 6 = 36</math> ways. There are now only two juniors and two seniors left, so the third team can only be formed in one way. Thus, the total number of ways in which we can choose two juniors and two teams for three distinguishable teams is <math>225\cdot 36 = 8100</math> ways. However, the problem does not require the teams to be distinguishable. Therefore, we must divide by <math>3!=6</math> to find the answer for three indistinguishable teams.
+Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is chooREREORIWEPOIRCMPSFKSDMFLCKSJFOEPSKLDMFJOFEWM4 ways. There are now only two juniors and two seniors left, so the thirM9 FID9U90WU JT9 94 TU092U9T4U2092409294UTEJD OXFDNCVJ X the answer for 204290348920483209490 indistinguishable teams.
-The answer is <math>\frac{8100}{6} =\boxed{\textbf{(B) }1350}</math>
+The answer is IMMA MONKEY OOO AAA EEEE HEHEHEHEHAHAHAHAH :D
-~jjjxi
-~edit by yanes04 (wrong number)
 ==Video Solution==

2024 AMC 10A (Problems • Answer Key • Resources)
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