AIME 2025 I · 第 3 题

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

Solution 3

Denote the number of people who chose strawberry, vanilla, or chocolate as (S, V, C). Then, as S < V < C, we just need to find values of S, V, and C such that S + V + C = 9. Notice S can only be 1 or 2 as S = 3 will result in V + C = 6 and it just won't work for S < V < C. So using these two values, we get that the possible triples of (S, V, C) are: (1, 3, 5), (2, 3, 4) and (1, 2, 6). Now, let's consider (S, V, C) = (1, 3, 5). If we start with the strawberry people, notice there are ${9\choose 1}$ possibilities. Now, we see there are 8 different people waiting to be assigned to the 3 vanilla people therefore there are ${8\choose 3}$ ways for this to work. We can now go down the list to get: ${9\choose 1}{8\choose 3}{5\choose 5} + {9\choose 2}{7\choose 3}{4\choose 4} + {9\choose 1}{8\choose 2}{6\choose 6}$ which gives a grand total of $2016$ possibilities. The remainder when $N$ is divided by $1000$ is $\boxed{016}$.

~ilikemath247365

Solution 4

We start by finding the only 3 possible cases, since $C>V>S$. We arrive at

$(6, 2, 1) = {9 \choose 6,2,1} = \frac{9 \cdot 8 \cdot 7 \cdot 6!}{6! \cdot 2! \cdot 1!} = 252$ $(5, 3, 1) = {9 \choose 5,3,1} = \frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{5! \cdot 3! \cdot 1!} = 504$ $(4, 3, 2) = {9 \choose 4,3,2} = \frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4!}{4! \cdot 3! \cdot 2!} = 1260$

Summing these up, we get $252+504+1260= 2\boxed{016}$

~shreyan.chethan, cleaned up by cweu001 and dareckolo

Solution 5

Let $c, v, s$ be chocolate, vanilla, and strawberry respectively. We will also use the notation $(s, v, c)$ to denote the number of chocolate, vanilla, and strawberry respectively. We have this condition, $c>v>s$. Now, we can list out the cases since there wont be a lot. First, we have:

$$(1,2,6) \Longrightarrow \binom{9}{1} \cdot \binom{8}{2} = 252$$ . Second, we have:

$$(1,3,5) \Longrightarrow \binom{9}{1} \cdot \binom{8}{3} = 504$$ , and last, we have:

$$(2,3,4) \Longrightarrow \binom{9}{2} \cdot \binom{7}{3} = 1260$$ . Summing them we get $2016$, then, we take $\mod 1000$, to get $\boxed{016}$

-jb2015007

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=J-0BapU4Yuk

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~HungryCalculator

Video Solution(Super Fun and Easy!)

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Video Solution by Mathletes Corner

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Video Solution by yjtest

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~yjtest