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Difference between revisions of "2024 AMC 10B Problems/Problem 20"

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==Solution 1==
 
==Solution 1==
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Let <math>A_R, A_L, B_R, B_L, C_R, C_L</math> denote the shoes.
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There are <math>6</math> ways to choose the first shoe. WLOG, assume it is <math>A_R</math>. We have <math>A_R,</math> __, __, __, __, __.
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<math>~~~~~</math> Case <math>1</math>: The next shoe in line is <math>A_L</math>. We have <math>A_R, A_L,</math> __, __, __, __. Now, the next shoe in line must be either <math>B_L</math> or <math>C_L</math>. There are <math>2</math> ways to choose which one, but assume WLOG that it is <math>B_L</math>. We have <math>A_R, A_L, B_L,</math> __, __, __.
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<math>~~~~~ ~~~~~</math> Subcase <math>1</math>: The next shoe in line is <math>B_R</math>. We have <math>A_R, A_L, B_L, B_R,</math> __, __. The only way to finish is <math>A_R, A_L, B_L, B_R, C_R, C_L</math>.
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<math>~~~~~ ~~~~~</math> Subcase <math>2</math>: The next shoe in line is <math>C_L</math>. We have <math>A_R, A_L, B_L, C_L,</math> __, __. The only way to finish is <math>A_R, A_L, B_L, C_L, C_R, B_R</math>.
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<math>~~~~~</math> In total, this case has <math>(6)(2)(1 + 1) = 24</math> orderings.
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<math>~~~~~</math> Case <math>2</math>: The next shoe in line is either <math>B_R</math> or <math>C_R</math>. There are <math>2</math> ways to choose which one, but assume WLOG that it is <math>B_R</math>. We have <math>A_R, B_R,</math> __, __, __, __.
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<math>~~~~~ ~~~~~</math> Subcase <math>1</math>: The next shoe is <math>B_L</math>. We have <math>A_R, B_R, B_L,</math> __, __, __.
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<math>~~~~~ ~~~~~ ~~~~~</math> Sub-subcase <math>1</math>: The next shoe in line is <math>A_L</math>. We have <math>A_R, B_R, B_L, A_L,</math> __, __. The only way to finish is <math>A_R, B_R, B_L, A_L, C_L, C_R</math>.
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<math>~~~~~ ~~~~~ ~~~~~</math> Sub-subcase <math>2</math>: The next shoe in line is <math>C_L</math>. We have <math>A_R, B_R, B_L, C_L,</math> __, __. The remaining shoes are <math>C_R</math> and <math>A_L</math>, but these shoes cannot be next to each other, so this sub-subcase is impossible.
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<math>~~~~~ ~~~~~</math> Subcase <math>2</math>: The next shoe is <math>C_R</math>. We have <math>A_R, B_R, C_R,</math> __, __, __. The next shoe in line must be <math>C_L</math>, so we have <math>A_R, B_R, C_R, C_L,</math> __, __. There are <math>2</math> ways to finish, which are <math>A_R, B_R, C_R, C_L, A_L, B_L</math> and <math>A_R, B_R, C_R, C_L, B_L, A_L</math>.
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<math>~~~~~</math> In total, this case has <math>(6)(2)(1 + 2) = 36</math> orderings.
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Our final answer is <math>24 + 36 = \boxed{\textbf{(A) } 60}</math>
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=B|num-b=19|num-a=21}}
 
{{AMC10 box|year=2024|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:00, 14 November 2024

Problem

Three different pairs of shoes are placed in a row so that no left shoe is next to a right shoe from a different pair. In how many ways can these six shoes be lined up?

$\textbf{(A) } 60 \qquad\textbf{(B) } 72 \qquad\textbf{(C) } 90 \qquad\textbf{(D) } 108 \qquad\textbf{(E) } 120$

Solution 1

Let $A_R, A_L, B_R, B_L, C_R, C_L$ denote the shoes.


There are $6$ ways to choose the first shoe. WLOG, assume it is $A_R$. We have $A_R,$ __, __, __, __, __.


$~~~~~$ Case $1$: The next shoe in line is $A_L$. We have $A_R, A_L,$ __, __, __, __. Now, the next shoe in line must be either $B_L$ or $C_L$. There are $2$ ways to choose which one, but assume WLOG that it is $B_L$. We have $A_R, A_L, B_L,$ __, __, __.


$~~~~~ ~~~~~$ Subcase $1$: The next shoe in line is $B_R$. We have $A_R, A_L, B_L, B_R,$ __, __. The only way to finish is $A_R, A_L, B_L, B_R, C_R, C_L$.


$~~~~~ ~~~~~$ Subcase $2$: The next shoe in line is $C_L$. We have $A_R, A_L, B_L, C_L,$ __, __. The only way to finish is $A_R, A_L, B_L, C_L, C_R, B_R$.


$~~~~~$ In total, this case has $(6)(2)(1 + 1) = 24$ orderings.


$~~~~~$ Case $2$: The next shoe in line is either $B_R$ or $C_R$. There are $2$ ways to choose which one, but assume WLOG that it is $B_R$. We have $A_R, B_R,$ __, __, __, __.


$~~~~~ ~~~~~$ Subcase $1$: The next shoe is $B_L$. We have $A_R, B_R, B_L,$ __, __, __.


$~~~~~ ~~~~~ ~~~~~$ Sub-subcase $1$: The next shoe in line is $A_L$. We have $A_R, B_R, B_L, A_L,$ __, __. The only way to finish is $A_R, B_R, B_L, A_L, C_L, C_R$.


$~~~~~ ~~~~~ ~~~~~$ Sub-subcase $2$: The next shoe in line is $C_L$. We have $A_R, B_R, B_L, C_L,$ __, __. The remaining shoes are $C_R$ and $A_L$, but these shoes cannot be next to each other, so this sub-subcase is impossible.


$~~~~~ ~~~~~$ Subcase $2$: The next shoe is $C_R$. We have $A_R, B_R, C_R,$ __, __, __. The next shoe in line must be $C_L$, so we have $A_R, B_R, C_R, C_L,$ __, __. There are $2$ ways to finish, which are $A_R, B_R, C_R, C_L, A_L, B_L$ and $A_R, B_R, C_R, C_L, B_L, A_L$.


$~~~~~$ In total, this case has $(6)(2)(1 + 2) = 36$ orderings.


Our final answer is $24 + 36 = \boxed{\textbf{(A) } 60}$

See also

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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