Difference between revisions of "2024 AMC 8 Problems/Problem 13"

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==Solution 2==
 
==Solution 2==
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Any combination can be written as some re-arrangement of <math>UUUDDD</math>. Clearly we must end going down, and start going up, so we need the number of ways to insert 2 <math>U</math>'s and 2 <math>D</math>'s into <math>U\, \_ \, \_ \, \_ \, \_ \, D</math>. There are <math>{4\choose 2}=6</math> ways, but we have to remove the case <math>UDDUUD</math>, giving us <math>\boxed{\textbf{(B)}\ 5}</math>.
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- We know there are no more cases since there will be at least one <math>U</math> before we have a <math>D</math> (from the first <math>U</math>), at least two <math>U</math>'S before two <math>D</math>'s (since we removed the one case), and at least three <math>U</math>'s before three <math>D</math>'s, as we end with the third <math>D</math>.
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~Sahan Wijetunga
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==Solution 3==
 
These numbers are clearly the Catalan numbers. Since we have 6 steps, we need the third Catalan number, which is <math>\boxed{\textbf{(B)}\ 5}</math>.
 
These numbers are clearly the Catalan numbers. Since we have 6 steps, we need the third Catalan number, which is <math>\boxed{\textbf{(B)}\ 5}</math>.
 
~andliu766
 
~andliu766

Revision as of 21:28, 26 January 2024

Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground? (For example, one sequence of hops is up-up-down-down-up-down.)

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$

Solution 1

Looking at the answer choices, you see that you can list them out. Doing this gets you:

UUDDUD

UDUDUD

UUUDDD

UDUUDD

UUDUDD

Counting all the paths listed above gets you 5 or B.

-ALWAYSRIGHT11

Solution 2

Any combination can be written as some re-arrangement of $UUUDDD$. Clearly we must end going down, and start going up, so we need the number of ways to insert 2 $U$'s and 2 $D$'s into $U\, \_ \, \_ \, \_ \, \_ \, D$. There are ${4\choose 2}=6$ ways, but we have to remove the case $UDDUUD$, giving us $\boxed{\textbf{(B)}\ 5}$.


- We know there are no more cases since there will be at least one $U$ before we have a $D$ (from the first $U$), at least two $U$'S before two $D$'s (since we removed the one case), and at least three $U$'s before three $D$'s, as we end with the third $D$.

~Sahan Wijetunga

Solution 3

These numbers are clearly the Catalan numbers. Since we have 6 steps, we need the third Catalan number, which is $\boxed{\textbf{(B)}\ 5}$. ~andliu766

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/X5Xk0wYXypk

Video Solution 2 by Math-X (First fully understand the problem!!!)

https://www.youtube.com/watch?v=Td6Z68YCuQw

~Math-X

Video Solution 3 by OmegaLearn.org

https://youtu.be/dM1wvr7mPQs

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=-kCN6R9U944

See Also

2024 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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 AJHSME/AMC 8 Problems and Solutions

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