Difference between revisions of "2024 AMC 10B Problems/Problem 17"
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For the case of a 5-way tie, we have <math>\dbinom{5}{5}=1</math> cases. We can assume this leads to an answer that ends in 1, leaving only <math>B</math> and <math>D</math>. By just accounting for the cases for a 1-way tie (<math>5!=120</math>) and 2-way tie (<math>\dbinom{5}{2}\cdot4\cdot3!=240</math>), it immediately shows that <math>B</math> does not work. Therefore, the answer is <math>\boxed{\text{(D) }431}</math> | For the case of a 5-way tie, we have <math>\dbinom{5}{5}=1</math> cases. We can assume this leads to an answer that ends in 1, leaving only <math>B</math> and <math>D</math>. By just accounting for the cases for a 1-way tie (<math>5!=120</math>) and 2-way tie (<math>\dbinom{5}{2}\cdot4\cdot3!=240</math>), it immediately shows that <math>B</math> does not work. Therefore, the answer is <math>\boxed{\text{(D) }431}</math> | ||
~shreyan.chethan | ~shreyan.chethan | ||
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(note that with 2 remaining choices the expected value of guessing is 3, and leaving it blank gives 1.5, so guessing after the 1 is deduced is statistically positive) | (note that with 2 remaining choices the expected value of guessing is 3, and leaving it blank gives 1.5, so guessing after the 1 is deduced is statistically positive) | ||
Revision as of 13:02, 15 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 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
Solution 3 (LAST EFFORT)
For the case of a 5-way tie, we have 
cases. We can assume this leads to an answer that ends in 1, leaving only 
and 
. By just accounting for the cases for a 1-way tie (
) and 2-way tie (
), it immediately shows that does not work. Therefore, the answer is 
~shreyan.chethan
(note that with 2 remaining choices the expected value of guessing is 3, and leaving it blank gives 1.5, so guessing after the 1 is deduced is statistically positive)
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 | |
| 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. 
