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

(→‎Video Solution 2 by SpreadTheMathLove)
(→‎Solution 2 (just had to))
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~EaZ_Shadow
 
~EaZ_Shadow
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==Solution 3(observation)==
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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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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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LLLRRR --> <math>3!\cdot 2!=12</math>
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RLLLRR --> <math>3!=6</math>
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RRLLLR --> <math>3!=6</math>
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RRRLLL --> <math>3!\cdot 2!=12.</math>
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LRRRLL --> <math>3!=6</math>
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LLRRRL --> <math>3!=6</math>
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LRRLLR --> <math>3!=6</math>
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RLLRRL --> <math>3!=6</math>
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We have <math>2\cdot 12+6\cdot 6=60.</math>
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~nevergonnagiveup
  
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

Revision as of 23:47, 16 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(observation)

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.

Below are the cases(note that once an L is fixed the R adjacent to it is also fixed due to the constraint): LLLRRR --> $3!\cdot 2!=12$ RLLLRR --> $3!=6$ RRLLLR --> $3!=6$ RRRLLL --> $3!\cdot 2!=12.$ LRRRLL --> $3!=6$ LLRRRL --> $3!=6$ LRRLLR --> $3!=6$ RLLRRL --> $3!=6$

We have $2\cdot 12+6\cdot 6=60.$

~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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