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

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<math>\textbf{(A) } 60 \qquad\textbf{(B) } 72 \qquad\textbf{(C) } 90 \qquad\textbf{(D) } 108 \qquad\textbf{(E) } 120</math>
 
<math>\textbf{(A) } 60 \qquad\textbf{(B) } 72 \qquad\textbf{(C) } 90 \qquad\textbf{(D) } 108 \qquad\textbf{(E) } 120</math>
  
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==Solution 1 (You can make changes or put your solution before mine if you have a better one)==
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==Solution 1==
  
 
Let <math>A_R, A_L, B_R, B_L, C_R, C_L</math> denote the shoes.
 
Let <math>A_R, A_L, B_R, B_L, C_R, C_L</math> denote the shoes.
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==Solution 3(focus on restrictions)==
 
==Solution 3(focus on restrictions)==
  
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Notice that you cannot have LRL or RLR, since you are guaranteed an R from a different pair from an L. This means you can either have three L's in a row, three R's in a row, or you have two R's between two L's and two L's between two R's.  
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Notice that you cannot have <math>LRL</math> or <math>RLR</math> in a row, since you are guaranteed an <math>R</math> and an <math>L</math> from a different pair. This means you can either have three <math>L</math>'s in a row, three <math>R</math>'s in a row, or you have two <math>R</math>'s between two <math>L</math>'s and two <math>L</math>'s between two <math>R</math>'s.  
  
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Below are the cases(note that once an L is fixed the R adjacent to it is also fixed due to the constraint):
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Below are the cases(note that once an <math>L</math> is fixed the <math>R</math> adjacent to it is also fixed due to the constraint):
 
\begin{align*}
 
\begin{align*}
 
LLLRRR \Rightarrow 3!\cdot 2!=12\\
 
LLLRRR \Rightarrow 3!\cdot 2!=12\\
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RLLLRR --> 3!=6\\
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RLLLRR \Rightarrow 3!=6\\
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RRLLLR --> 3!=6\\
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RRLLLR \Rightarrow 3!=6\\
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RRRLLL --> 3!\cdot 2!=12\\
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RRRLLL \Rightarrow 3!\cdot 2!=12\\
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LRRRLL --> 3!=6\\
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LRRRLL \Rightarrow 3!=6\\
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LLRRRL --> 3!=6\\
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LLRRRL \Rightarrow 3!=6\\
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LRRLLR --> 3!=6\\
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LRRLLR \Rightarrow 3!=6\\
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RLLRRL --> 3!=6\\
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RLLRRL \Rightarrow 3!=6\\
 
\end{align*}
 
\end{align*}
  
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We have <math>2\cdot 12+6\cdot 6=60.</math>
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We have <math>2\cdot 12+6\cdot 6=\boxed{\textbf{(A) }60}.</math>
  
 
~nevergonnagiveup
 
~nevergonnagiveup

Latest revision as of 20:46, 22 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 $24 + 36 = \boxed{\textbf{(A) } 60}$

Solution 2 (just had to)

Alright so first off, an obvious configuration is $LLLRRR$, where I will not leave distinction between the L’s or the R’s to simplify things. This has $3!$ ways to range the $L$’s and $2!$ ways to arrange the $R$’s, or 12 ways in total. Notice that we can reverse, the order into $RRRLLL$, which I will be do many times, yields a total of 24. Now, trying out some cases, we find that $RLLRRL$, works, so there are $6$ ways to arrange the pairs of $RL$ and $2$ ways to choose the orientation of one pair (which determines the other pairs’ orientation), yielding a total of 12 ways. Lastly, we can have $RLLLRR$, which has $3!$ ways to determine the $L$’s which determine the $R$’s. Notice that we can change the R’s to L’s and vice versa, or the configuration $LRRRLL$. We can also flip the ordering to get $RRLLLR$ and $LLRRRL$. This case yields $6\cdot 2 \cdot 2$ or $24$ ways. Adding the cases up, we get $60$ as our answer, or $\boxed{A}$.

~EaZ_Shadow


Solution 3(focus on restrictions)

Notice that you cannot have $LRL$ or $RLR$ in a row, since you are guaranteed an $R$ and an $L$ from a different pair. This means you can either have three $L$'s in a row, three $R$'s in a row, or you have two $R$'s between two $L$'s and two $L$'s between two $R$'s.

Below are the cases(note that once an $L$ is fixed the $R$ adjacent to it is also fixed due to the constraint): \begin{align*} LLLRRR \Rightarrow 3!\cdot 2!=12\\ RLLLRR \Rightarrow 3!=6\\ RRLLLR \Rightarrow 3!=6\\ RRRLLL \Rightarrow 3!\cdot 2!=12\\ LRRRLL \Rightarrow 3!=6\\ LLRRRL \Rightarrow 3!=6\\ LRRLLR \Rightarrow 3!=6\\ RLLRRL \Rightarrow 3!=6\\ \end{align*}

We have $2\cdot 12+6\cdot 6=\boxed{\textbf{(A) }60}.$

~nevergonnagiveup

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=yYpnHoTQNi4

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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