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Difference between revisions of "2024 AMC 10B Problems/Problem 17"
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<math>3</math> snails tied: We have <math>ABC, D, E</math>. There are <math>3! = 6</math> ways to determine the ranking of the <math>3</math> groups. There are <math>5\choose2</math> ways to determine the two snails not involved in the tie. So <math>6 \cdot 10 = 60</math> ways.
 
<math>3</math> snails tied: We have <math>ABC, D, E</math>. There are <math>3! = 6</math> ways to determine the ranking of the <math>3</math> groups. There are <math>5\choose2</math> ways to determine the two snails not involved in the tie. So <math>6 \cdot 10 = 60</math> ways.
  
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<math>2</math> snails tied: We have <math>AB, C, D, E</math>. There are <math>4! = 24</math> ways to determine the ranking of the <math>4</math> groups. There are <math>5\choose{3}</math> ways to determine the three snail not involved in the tie. So <math>24 \cdot 10 = 240</math> ways.
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<math>2</math> snails tied: We have <math>AB, C, D, E</math>. There are <math>4! = 24</math> ways to determine the ranking of the <math>4</math> groups. There are <math>5\choose{3}</math> ways to determine the three snails not involved in the tie. So <math>24 \cdot 10 = 240</math> ways.
  
 
It's impossible to have "1 snail tie", so that case has <math>0</math> ways.
 
It's impossible to have "1 snail tie", so that case has <math>0</math> ways.
  
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Finally, there are no ties. We just arrange the <math>5</math> snail, so <math>5! = 120</math> ways.
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Finally, there are no ties. We just arrange the <math>5</math> snails, so <math>5! = 120</math> ways.
  
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The answer is <math>1+10+60+240+0+120 = \boxed{431}</math>.
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The answer is <math>1+10+60+240+0+120 = \boxed{\text{(D) }431}</math>.
  
 
~lprado
 
~lprado
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The answer is <math>120+240+60+10+1=\boxed{\text{(D) }431}</math> ~Tacos_are_yummy_1
 
The answer is <math>120+240+60+10+1=\boxed{\text{(D) }431}</math> ~Tacos_are_yummy_1
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==Solution 3 (Get Lucky)==
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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>
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~shreyan.chethan
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Notice that with 2 remaining choices, through guessing, the expected value of your points is <math>\frac{6}{2}=3</math>, and not answering gives 1.5 points. Therefore, you gain an expected value of <math>1.5</math> points by answering. It would be best to always guess whenever you can get rid of <math>2</math> or more answers. If you can only eliminate <math>1</math> answer, not guessing will give you <math>1.5</math> points, but guessing also gives you an expected value of <math>1.5</math> points. I usually leave it blank if I can only get rid of <math>1</math> answer choice.
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~shreyan.chethan, edited by BenjaminDong01, last part added by PerseverePlayer
  
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
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==Video Solution 2 by SpreadTheMathLove==
 
==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
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https://www.youtube.com/watch?v=Q7fwWZ89MC8
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2024|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:39, 23 November 2024

Problem

In a race among $5$ 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?

$\textbf{(A) } 180 \qquad\textbf{(B) } 361 \qquad\textbf{(C) } 420 \qquad\textbf{(D) } 431 \qquad\textbf{(E) } 720$

Solution 1

We perform casework based on how many snails tie. Let's say we're dealing with the following snails: $A,B,C,D,E$.

$5$ snails tied: All $5$ snails tied for $1$st place, so only $1$ way.

$4$ snails tied: $A,B,C,D$ all tied, and $E$ either got $1$st or last. ${5}\choose{1}$ ways to choose who isn't involved in the tie and $2$ ways to choose if that snail gets first or last, so $10$ ways.

$3$ snails tied: We have $ABC, D, E$. There are $3! = 6$ ways to determine the ranking of the $3$ groups. There are $5\choose2$ ways to determine the two snails not involved in the tie. So $6 \cdot 10 = 60$ ways.

$2$ snails tied: We have $AB, C, D, E$. There are $4! = 24$ ways to determine the ranking of the $4$ groups. There are $5\choose{3}$ ways to determine the three snails not involved in the tie. So $24 \cdot 10 = 240$ ways.

It's impossible to have "1 snail tie", so that case has $0$ ways.

Finally, there are no ties. We just arrange the $5$ snails, so $5! = 120$ ways.

The answer is $1+10+60+240+0+120 = \boxed{\text{(D) }431}$.

~lprado

Solution 2 (Solution 1 but less words)

Split the problem into cases. A tie of $n$ snails has $\dbinom{5}{n}$ ways to choose the snails that are tied, $6-n$ ways to choose which place they tie for, and $(5-n)!$ to place the remaining snails.

1. No tie $\implies5!=120$

2. Tie of 2 snails $\implies\dbinom{5}{2}\cdot4\cdot3!=240$

3. Tie of 3 snails $\implies\dbinom{5}{3}\cdot3\cdot2!=60$

4. Tie of 4 snails $\implies\dbinom{5}{4}\cdot2=10$

5. Tie of all 5 snails $\implies1$

The answer is $120+240+60+10+1=\boxed{\text{(D) }431}$ ~Tacos_are_yummy_1

Solution 3 (Get Lucky)

For the case of a 5-way tie, we have $\dbinom{5}{5}=1$ cases. We can assume this leads to an answer that ends in 1, leaving only $B$ and $D$. By just accounting for the cases for a 1-way tie ($5!=120$) and 2-way tie ($\dbinom{5}{2}\cdot4\cdot3!=240$), it immediately shows that $B$ does not work. Therefore, the answer is $\boxed{\text{(D) }431}$ ~shreyan.chethan

Notice that with 2 remaining choices, through guessing, the expected value of your points is $\frac{6}{2}=3$, and not answering gives 1.5 points. Therefore, you gain an expected value of $1.5$ points by answering. It would be best to always guess whenever you can get rid of $2$ or more answers. If you can only eliminate $1$ answer, not guessing will give you $1.5$ points, but guessing also gives you an expected value of $1.5$ points. I usually leave it blank if I can only get rid of $1$ answer choice. ~shreyan.chethan, edited by BenjaminDong01, last part added by PerseverePlayer

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)
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Problem 16 		Followed by
Problem 18
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