2024 AMC 10B Problems/Problem 17
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[hide]Problem
In a race among snails, there is at most one tie, but that tie can involve any number of snails. For example, the result might be that Dazzler is first; Abby, Cyrus, and Elroy are tied for second; and Bruna is fifth. How many different results of the race are possible?
Solution 1
We perform casework based on how many snails tie. Let's say we're dealing with the following snails: .
snails tied: All snails tied for st place, so only way.
snails tied: all tied, and either got st or last. ways to choose who isn't involved in the tie and ways to choose if that snail gets first or last, so ways.
snails tied: We have . There are ways to determine the ranking of the groups. There are ways to determine the two snails not involved in the tie. So ways.
snails tied: We have . There are ways to determine the ranking of the groups. There are ways to determine the three snails not involved in the tie. So ways.
It's impossible to have "1 snail tie", so that case has ways.
Finally, there are no ties. We just arrange the snails, so ways.
The answer is .
~lprado
Solution 2 (Solution 1 but less words)
Split the problem into cases. A tie of snails has ways to choose the snails that are tied, ways to choose which place they tie for, and to place the remaining snails.
1. No tie
2. Tie of 2 snails
3. Tie of 3 snails
4. Tie of 4 snails
5. Tie of all 5 snails
The answer is ~Tacos_are_yummy_1
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/c6nhclB5V1w?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=Q7fwWZ89MC8
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AMC 10 Problems and Solutions |
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