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

(Solution 2 (just had to))
(Solution 3(observation))
Line 52: Line 52:
  
  
==Solution 3(observation)==
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==Solution 3(focus on restrictions)==
  
 
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.  
 
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):
 
Below are the cases(note that once an L is fixed the R adjacent to it is also fixed due to the constraint):
LLLRRR --> <math>3!\cdot 2!=12</math>
+
\begin{align*}
RLLLRR --> <math>3!=6</math>
+
LLLRRR \Rightarrow 3!\cdot 2!=12\
RRLLLR --> <math>3!=6</math>
+
RLLLRR --> 3!=6\
RRRLLL --> <math>3!\cdot 2!=12.</math>
+
RRLLLR --> 3!=6\
LRRRLL --> <math>3!=6</math>
+
RRRLLL --> 3!\cdot 2!=12\
LLRRRL --> <math>3!=6</math>
+
LRRRLL --> 3!=6\
LRRLLR --> <math>3!=6</math>
+
LLRRRL --> 3!=6\
RLLRRL --> <math>3!=6</math>
+
LRRLLR --> 3!=6\
 +
RLLRRL --> 3!=6\
 +
\end{align*}
  
 
We have <math>2\cdot 12+6\cdot 6=60.</math>
 
We have <math>2\cdot 12+6\cdot 6=60.</math>

Revision as of 23:51, 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(focus on restrictions)

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): LLLRRR3!2!=12RLLLRR>3!=6RRLLLR>3!=6RRRLLL>3!2!=12LRRRLL>3!=6LLRRRL>3!=6LRRLLR>3!=6RLLRRL>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 (ProblemsAnswer KeyResources)
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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