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

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<math>UUDUDD</math>
 
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Counting all the paths listed above gets you <math>5</math> or B.
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Counting all the paths listed above gets you <math>5</math> or <math>B</math>.
  
 
-ALWAYSRIGHT11
 
-ALWAYSRIGHT11
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\\
 
-vockey(minor latex changes)
 
-vockey(minor latex changes)
  

Revision as of 16:01, 28 January 2024

Problem

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.)

2024-AMC8-q13.png

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

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 \\ -vockey(minor latex changes)

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

Solution 4

First step is U, last step is D.

After third step we can get only positions 3 or 1.

In the first case there is only one way UUUDDD.

In the second case we have two way to get this position UDU and UUD.

Similarly, we have two way return to position 0 (UDD and DUD).

Totally we have $1 + 2 \cdot 2 = \boxed{\textbf{(B)}\ 5}$.

vladimir.shelomovskii@gmail.com, vvsss

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 NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths


Video Solution by CosineMethod [🔥Fast and Easy🔥]

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

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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