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

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Our final answer is <math>24 + 36 = \boxed{\textbf{(A) } 60}</math>
 
Our final answer is <math>24 + 36 = \boxed{\textbf{(A) } 60}</math>
  
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==Solution 2 (just had to)==
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==Solution 2==
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Alright so first off, an obvious configuration is <math>LLLRRR</math>, where I will not leave distinction between the L’s or the R’s to simplify things. This has <math>3!</math> ways to range the <math>L</math>’s and <math>2!</math> ways to arrange the <math>R</math>’s, or 12 ways in total. Notice that we can reverse,  the order into <math>RRRLLL</math>, which I will be do many times, yields a total of 24. Now, trying out some cases, we find that <math>RLLRRL</math>, works, so there are <math>6</math> ways to arrange the pairs of <math>RL</math> and <math>2</math> ways to choose the orientation of one pair (which determines the other pairs’ orientation), yielding a total of 12 ways. Lastly, we can have <math>RLLLRR</math>, which has <math>3!</math> ways to determine the <math>L</math>’s which determine the <math>R</math>’s. Notice that we can change the R’s to L’s and vice versa, or the configuration <math>LRRRLL</math>. We can also flip the ordering to get <math>RRLLLR</math> and <math>LLRRRL</math>. This case yields <math>6\cdot 2 \cdot 2</math> or <math>24</math> ways. Adding the cases up, we get <math>60</math> as our answer, or <math>\boxed{A}</math>.
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We have two main cases:
  
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~EaZ_Shadow
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'''Case 1:''' LLLRRR or RRRLLL
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There are <math>3! = 6</math> ways to assign the shoes to the ordering. For now we'll focus on the <math>LLLRRR</math> Case and multiply by <math>2</math> since they're symmetrical. WLOG say we have <math>A_LB_LC_LC_RA_RB_R</math>. We can flip <math>A_L</math> and <math>B_L</math> and we can flip <math>A_R</math> and <math>B_R,</math> so we fet <math>6 \cdot 2 \cdot 2 = 24.</math> Multiplying by <math>2</math> for symmetry we get <math>48</math> ways for this case.
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'''Case 2:''' RRLLLR or RLLLRR
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Again, we'll focus on the <math>RRLLLR</math> Case and multiply by <math>2</math> since they're symmetrical. WLOG say we have <math>B_RA_RA_LB_LC_LC_R</math>. We can assign the shoes to the ordering in <math>6</math> ways. <math>B_R</math> can also be moved over to be next to <math>C_R</math>. Then we would have <math>A_RA_LB_LC_LC_RB_R</math> which is also a valid ordering. So this case gives us <math>6 \cdot 2 = 12</math> ways.
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The answer is <math>48 + 12 = \boxed{60}.</math>
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~[[User:grogg007|grogg007]], ~EaZ_Shadow
  
 
==Solution 3(focus on restrictions)==
 
==Solution 3(focus on restrictions)==

Revision as of 13:17, 25 August 2025

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

We have two main cases:

Case 1: LLLRRR or RRRLLL

There are $3! = 6$ ways to assign the shoes to the ordering. For now we'll focus on the $LLLRRR$ Case and multiply by $2$ since they're symmetrical. WLOG say we have $A_LB_LC_LC_RA_RB_R$. We can flip $A_L$ and $B_L$ and we can flip $A_R$ and $B_R,$ so we fet $6 \cdot 2 \cdot 2 = 24.$ Multiplying by $2$ for symmetry we get $48$ ways for this case.


Case 2: RRLLLR or RLLLRR

Again, we'll focus on the $RRLLLR$ Case and multiply by $2$ since they're symmetrical. WLOG say we have $B_RA_RA_LB_LC_LC_R$. We can assign the shoes to the ordering in $6$ ways. $B_R$ can also be moved over to be next to $C_R$. Then we would have $A_RA_LB_LC_LC_RB_R$ which is also a valid ordering. So this case gives us $6 \cdot 2 = 12$ ways.

The answer is $48 + 12 = \boxed{60}.$

~grogg007, ~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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