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Difference between revisions of "2024 AMC 10B Problems/Problem 17"

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==Problem==
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In a race among <math>5</math> 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?
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<math>\textbf{(A) } 180 \qquad\textbf{(B) } 361 \qquad\textbf{(C) } 420 \qquad\textbf{(D) } 431 \qquad\textbf{(E) } 720</math>
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==Solution 1==
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We perform casework based on how many snails tie. Let's say we're dealing with the following snails: <math>A,B,C,D,E</math>.
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<math>5</math> snails tied: All <math>5</math> snails tied for <math>1</math>st place, so only <math>1</math> way.
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<math>4</math> snails tied: <math>A,B,C,D</math> all tied, and <math>E</math> either got <math>1</math>st or last. <math>{5}\choose{1}</math> ways to choose who isn't involved in the tie and <math>2</math> ways to choose if that snail gets first or last, so <math>10</math> ways.
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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.
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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.
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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> snails, so <math>5! = 120</math> ways.
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The answer is <math>1+10+60+240+0+120 = \boxed{\text{(D) }431}</math>.
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~lprado
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==Solution 2 (Solution 1 but less words)==
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Split the problem into cases. A tie of <math>n</math> snails has <math>\dbinom{5}{n}</math> ways to choose the snails that are tied, <math>6-n</math> ways to choose which place they tie for, and <math>(5-n)!</math> to place the remaining snails.
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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>
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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>
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5. Tie of all 5 snails <math>\implies1</math>
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The answer is <math>120+240+60+10+1=\boxed{\text{(D) }431}</math> ~Tacos_are_yummy_1
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==Solution 3 (LAST EFFORT)==
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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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(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)
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==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
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https://youtu.be/c6nhclB5V1w?feature=shared
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~ Pi Academy
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=Q7fwWZ89MC8
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==See also==
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{{AMC10 box|year=2024|ab=B|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 13:02, 15 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 (LAST EFFORT)

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

(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
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All AMC 10 Problems and Solutions

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