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

Latest revision as of 01:01, 18 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 (You can make changes or put your solution before mine if you have a better one)

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

@@ Line 6: / Line 6: @@
 <math>\textbf{(A) } 60 \qquad\textbf{(B) } 72 \qquad\textbf{(C) } 90 \qquad\textbf{(D) } 108 \qquad\textbf{(E) } 120</math>
-==Solution 1 (Feel free to make changes or put your solution before mine if you have a better one)==
+==Solution 1 (You can make changes or put your solution before mine if you have a better one)==
 Let <math>A_R, A_L, B_R, B_L, C_R, C_L</math> denote the shoes.
@@ Line 45: / Line 45: @@
 Our final answer is <math>24 + 36 = \boxed{\textbf{(A) } 60}</math>
+==Solution 2 (just had to)==
+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>.
+~EaZ_Shadow
+==Solution 3(focus on restrictions)==
+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.
+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*}
+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 <math>2\cdot 12+6\cdot 6=\boxed{\textbf{(A) }60}.</math>
+~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==
 {{AMC10 box|year=2024|ab=B|num-b=19|num-a=21}}
 {{MAA Notice}}

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
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