Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (just had to))
m (→‎Solution 3(focus on restrictions))
Line 70: Line 70:
  
 
~nevergonnagiveup
 
~nevergonnagiveup
 +
 +
==Solution 4 (only two cases)==
 +
 +
We have two main cases:
 +
 +
'''Case 1:''' <math>LLLRRR or RRRLLL</math>
 +
 +
For now we'll focus on the <math>LLLRRR</math> case and multiply by <math>2</math> at the end since they're symmetrical. WLOG say we have <math>A_LB_LC_LC_RA_RB_R.</math> There are <math>3! = 6</math> ways to assign colors to the shoes. We can rearrange <math>A_L</math> and <math>B_L</math> in <math>2</math> ways and <math>A_R and B_R</math> in <math>2</math> ways. This case gives us <math>6 \cdot 2 \cdot 2 \cdot 2 = 48</math> ways
 +
 +
'''Case 2:''' <math>RRLLLR</math> or <math>RLLLRR</math>
 +
 +
Again, we'll focus on the <math>RRLLLR</math> case and multiply by <math>2.</math> WLOG say we have <math>A_RB_RA_LB_LC_LC_R.</math> There are <math>3! = 6</math> ways to assign colors to the shoes. Everything else is fixed. Multiplying by <math>2</math> we get <math>12</math> for this case.
 +
 +
The answer is <math>48 + 12 = \boxed{60}.</math>
 +
 +
~[[User:grogg007]]
  
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

Revision as of 10:39, 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 (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

Solution 4 (only two cases)

We have two main cases:

Case 1: $LLLRRR or RRRLLL$

For now we'll focus on the $LLLRRR$ case and multiply by $2$ at the end since they're symmetrical. WLOG say we have $A_LB_LC_LC_RA_RB_R.$ There are $3! = 6$ ways to assign colors to the shoes. We can rearrange $A_L$ and $B_L$ in $2$ ways and $A_R and B_R$ in $2$ ways. This case gives us $6 \cdot 2 \cdot 2 \cdot 2 = 48$ ways

Case 2: $RRLLLR$ or $RLLLRR$

Again, we'll focus on the $RRLLLR$ case and multiply by $2.$ WLOG say we have $A_RB_RA_LB_LC_LC_R.$ There are $3! = 6$ ways to assign colors to the shoes. Everything else is fixed. Multiplying by $2$ we get $12$ for this case.

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

~User:grogg007

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10B_Problems/Problem_20&oldid=258821"