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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==
 
==Solution 1==
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We perform casework based on how many people tie. Let's say we're dealing with the following people: <math>A,B,C,D,E</math>.
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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> people tied: All <math>5</math> people tied for <math>1</math>st place, so only <math>1</math> way.
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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> people 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 person gets first or last, so <math>10</math> ways.
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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> people 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 people not involved in the tie. So <math>6 \cdot 10 = 60</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> people 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 people 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 snail not involved in the tie. So <math>24 \cdot 10 = 240</math> ways.
  
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It's impossible to have "1 person tie", so that case has <math>0</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> people, so <math>5! = 120</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.
  
 
The answer is <math>1+10+60+240+0+120 = \boxed{431}</math>.
 
The answer is <math>1+10+60+240+0+120 = \boxed{431}</math>.
  
 
~lprado
 
~lprado
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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 01:30, 14 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 snail 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 $5$ snail, so $5! = 120$ ways.

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

~lprado

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

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