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Difference between revisions of "2025 AIME I Problems/Problem 3"

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==Solution 1==
 
==Solution 1==
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Let <math>c</math> be the number of players who choose chocolate, <math>v</math> be the number of players who choose vanilla, and <math>s</math> be the number of players who choose strawberry ice cream. We are given two pieces of information <math>c,v,s\ge 1</math> and <math>c+v+s=9.</math>
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By inspection the only solutions for <math>(c,v,s)</math> are <math>(2,3,4),(1,2,6),(1,3,5).</math>
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 +
Now we must choose which player chooses which flavor. For the general case <math>(c,v,s),</math> we begin by choose <math>c</math> of the <math>9</math> players who eat chocolate, then we choose <math>v</math> of the <math>9-c</math> players who vanilla, after this the amount of players who eat strawberry is fixed. Therefore the general formula is <math>\binom{9}{c}\binom{9-c}{v}.</math>
 +
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Therefore our final answer is, <cmath>\binom{9}{2}\binom{7}{3}+\binom{9}{1}\binom{8}{2}+\binom{9}{1}\binom{8}{3}=2\boxed{016}.</cmath>
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~ [[User:Mathkiddus|mathkiddus]]
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==Solution 2==
  
 
We apply casework on the scoops the team gets.
 
We apply casework on the scoops the team gets.
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~ [[User:zhenghua|zhenghua]]
 
~ [[User:zhenghua|zhenghua]]
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==Solution 2==
 
−
Let <math>c</math> be the number of players who choose chocolate, <math>v</math> be the number of players who choose vanilla, and <math>s</math> be the number of players who choose strawberry ice cream. We are given two pieces of information <math>c,v,s\ge 1</math> and <math>c+v+s=9.</math>
 
−
 
−
By inspection the only solutions for <math>(c,v,s)</math> are <math>(2,3,4),(1,2,6),(1,3,5).</math>
 
−
 
−
Now we must choose which player chooses which flavor. For the general case <math>(c,v,s),</math> we begin by choose <math>c</math> of the <math>9</math> players who eat chocolate, then we choose <math>v</math> of the <math>9-c</math> players who vanilla, after this the amount of players who eat strawberry is fixed. Therefore the general formula is <math>\binom{9}{c}\binom{9-c}{v}.</math>
 
−
 
−
Therefore our final answer is, <cmath>\binom{9}{2}\binom{7}{3}+\binom{9}{1}\binom{8}{2}+\binom{9}{1}\binom{8}{3}=2\boxed{016}.</cmath>
 
−
 
−
~ [[User:Mathkiddus|mathkiddus]]
 
 
==See also==
 
==See also==
 
{{AIME box|year=2025|num-b=2|num-a=4|n=I}}
 
{{AIME box|year=2025|num-b=2|num-a=4|n=I}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:46, 13 February 2025

Problem

The $9$ members of a baseball team went to an ice-cream parlor after their game. Each player had a single scoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was greater than the number of players who chose strawberry. Let $N$ be the number of different assignments of flavors to players that meet these conditions. Find the remainder when $N$ is divided by $1000.$

Solution 1

Let $c$ be the number of players who choose chocolate, $v$ be the number of players who choose vanilla, and $s$ be the number of players who choose strawberry ice cream. We are given two pieces of information $c,v,s\ge 1$ and $c+v+s=9.$

By inspection the only solutions for $(c,v,s)$ are $(2,3,4),(1,2,6),(1,3,5).$

Now we must choose which player chooses which flavor. For the general case $(c,v,s),$ we begin by choose $c$ of the $9$ players who eat chocolate, then we choose $v$ of the $9-c$ players who vanilla, after this the amount of players who eat strawberry is fixed. Therefore the general formula is $\binom{9}{c}\binom{9-c}{v}.$

Therefore our final answer is, \[\binom{9}{2}\binom{7}{3}+\binom{9}{1}\binom{8}{2}+\binom{9}{1}\binom{8}{3}=2\boxed{016}.\]

~ mathkiddus

Solution 2

We apply casework on the scoops the team gets.

Case 1: The scoops are $6,2,1$. Then we have $\binom{9}{6}\cdot \binom{3}{2} = 252$.

Case 2: The scoops are $5,3,1$. Then we have $\binom{9}{5}\cdot \binom{4}{3} = 504$.

Case 3: The scoops are $4,3,2$. Then we have $\binom{9}{4}\cdot \binom{5}{3} = 1260$.

Thus the answer is $252+504+1260=2\boxed{016}$.

~ zhenghua

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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