Difference between revisions of "2024 AMC 10B Problems/Problem 20"
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− | Notice that you cannot have LRL or RLR, since you are guaranteed an R from a different pair | + | 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 L is fixed the R adjacent to it is also fixed due to the constraint): | + | 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*} | \begin{align*} | ||
LLLRRR \Rightarrow 3!\cdot 2!=12\ | LLLRRR \Rightarrow 3!\cdot 2!=12\ | ||
− | RLLLRR | + | RLLLRR \Rightarrow 3!=6\ |
− | RRLLLR | + | RRLLLR \Rightarrow 3!=6\ |
− | RRRLLL | + | RRRLLL \Rightarrow 3!\cdot 2!=12\ |
− | LRRRLL | + | LRRRLL \Rightarrow 3!=6\ |
− | LLRRRL | + | LLRRRL \Rightarrow 3!=6\ |
− | LRRLLR | + | LRRLLR \Rightarrow 3!=6\ |
− | RLLRRL | + | RLLRRL \Rightarrow 3!=6\ |
\end{align*} | \end{align*} | ||
− | We have <math>2\cdot 12+6\cdot 6=60.</math> | + | We have <math>2\cdot 12+6\cdot 6=\boxed{\textbf{(A) }60}.</math> |
~nevergonnagiveup | ~nevergonnagiveup |
Revision as of 01:01, 18 November 2024
Contents
[hide]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?
Solution 1 (You can make changes or put your solution before mine if you have a better one)
Let denote the shoes.
There are ways to choose the first shoe. WLOG, assume it is . We have __, __, __, __, __.
Case : The next shoe in line is . We have __, __, __, __. Now, the next shoe in line must be either or . There are ways to choose which one, but assume WLOG that it is . We have __, __, __.
Subcase : The next shoe in line is . We have __, __. The only way to finish is .
Subcase : The next shoe in line is . We have __, __. The only way to finish is .
In total, this case has orderings.
Case : The next shoe in line is either or . There are ways to choose which one, but assume WLOG that it is . We have __, __, __, __.
Subcase : The next shoe is . We have __, __, __.
Sub-subcase : The next shoe in line is . We have __, __. The only way to finish is .
Sub-subcase : The next shoe in line is . We have __, __. The remaining shoes are and , but these shoes cannot be next to each other, so this sub-subcase is impossible.
Subcase : The next shoe is . We have __, __, __. The next shoe in line must be , so we have __, __. There are ways to finish, which are and .
In total, this case has orderings.
Our final answer is
Solution 2 (just had to)
Alright so first off, an obvious configuration is , where I will not leave distinction between the L’s or the R’s to simplify things. This has ways to range the ’s and ways to arrange the ’s, or 12 ways in total. Notice that we can reverse, the order into , which I will be do many times, yields a total of 24. Now, trying out some cases, we find that , works, so there are ways to arrange the pairs of and ways to choose the orientation of one pair (which determines the other pairs’ orientation), yielding a total of 12 ways. Lastly, we can have , which has ways to determine the ’s which determine the ’s. Notice that we can change the R’s to L’s and vice versa, or the configuration . We can also flip the ordering to get and . This case yields or ways. Adding the cases up, we get as our answer, or .
~EaZ_Shadow
Solution 3(focus on restrictions)
Notice that you cannot have or in a row, since you are guaranteed an and an from a different pair. This means you can either have three 's in a row, three 's in a row, or you have two 's between two 's and two 's between two 's.
Below are the cases(note that once an is fixed the adjacent to it is also fixed due to the constraint):
We have
~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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