Difference between revisions of "2024 AMC 10B Problems/Problem 20"
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Our final answer is <math>24 + 36 = \boxed{\textbf{(A) } 60}</math> | Our final answer is <math>24 + 36 = \boxed{\textbf{(A) } 60}</math> | ||
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+ | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ||
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+ | https://youtu.be/c6nhclB5V1w?feature=shared | ||
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+ | ~ Pi Academy | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2024|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:05, 14 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 (Feel free to 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
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/c6nhclB5V1w?feature=shared
~ Pi Academy
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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