Difference between revisions of "2024 AMC 10B Problems/Problem 17"
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1. No tie <math>\implies5!=120</math> | 1. No tie <math>\implies5!=120</math> | ||
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2. Tie of 2 snails <math>\implies\dbinom{5}{2}\cdot4\cdot3!=240</math> | 2. Tie of 2 snails <math>\implies\dbinom{5}{2}\cdot4\cdot3!=240</math> | ||
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3. Tie of 3 snails <math>\implies\dbinom{5}{3}\cdot3\cdot2!=60</math> | 3. Tie of 3 snails <math>\implies\dbinom{5}{3}\cdot3\cdot2!=60</math> | ||
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4. Tie of 4 snails <math>\implies\dbinom{5}{4}\cdot2=10</math> | 4. Tie of 4 snails <math>\implies\dbinom{5}{4}\cdot2=10</math> | ||
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5. Tie of all 5 snails <math>\implies1</math> | 5. Tie of all 5 snails <math>\implies1</math> | ||
Revision as of 12:07, 14 November 2024
Contents
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 snail 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 snail, 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
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
