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

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==Problem==
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== 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 <math>6</math> hops, and end up back on the ground?
 
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 <math>6</math> hops, and end up back on the ground?
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==Solution 4==
 
==Solution 4==
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First step is U, last step is D.
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First step must be a U, last step must be a D.
  
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After third step we can get only positions 3 or 1.
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After third step we can get only position 3 or position 1.
  
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In the first case there is only one way UUUDDD.
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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.
 
In the second case we have two way to get this position UDU and UUD.
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
  
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https://www.youtube.com/watch?v=-kCN6R9U944
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https://www.youtube.com/watch?v=-Yummy
 +
 
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 +
 
https://youtu.be/ktzijuZtDas&t=1238
 
https://youtu.be/ktzijuZtDas&t=1238
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 +
==Video Solution by Dr. David==
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https://youtu.be/r3FtOOYEces
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 +
==Video Solution by WhyMath==
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https://youtu.be/6Bg0Z0jcInw
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2024|num-b=12|num-a=14}}
 
{{AMC8 box|year=2024|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:58, 20 November 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:

$\mathit{UUDDUD}$

$\mathit{UDUDUD}$

$\mathit{UUUDDD}$

$\mathit{UDUUDD}$

$\mathit{UUDUDD}$

Counting all the paths listed above gets you $\boxed{\textbf{(B)} \ 5}$.

~ALWAYSRIGHT11 ~vockey(minor edits) ~Johnxyz1 (use mathit for better letter space)

Solution 2

Any combination can be written as some re-arrangement of $\mathit{UUUDDD}$. 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 $\mathit{UDDUUD}$, 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 must be a U, last step must be a D.

After third step we can get only position 3 or position 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).

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

vladimir.shelomovskii@gmail.com, vvsss

Solution 5 (Complementary Counting)

We can find the total cases then deduct the ones that don't work.

Let $U$ represent "Up" and $D$ represent "Down". We know that in order to land back at the bottom of the stairs, we must have an equal number of $U$'s and $D$'s, therefore six hops means $3$ of each.

The number of ways to arrange $3$ $U$'s and $3$ $D$'s is $\dfrac{6!}{(3!)^2}=\dfrac{720}{36}=20$.

Case $1$: Start with $\mathit D$

Case $2$: Start with $\mathit{UDD}$

Case $3$: Start with $\mathit{UUDDD}$

Case $4$: Start with $\mathit{UDUDD}$

Case $1$ is asking us how many ways there are to arrange $3$ $U$'s and $2$ $D$'s, which is $\dfrac{5!}{3!2!}=\dfrac{120}{12}=10$.

Case $2$ is asking us how many ways there are to arrange $2$ $U$'s and $1$ $D$, which is $\dfrac{3!}{2!1!}=\dfrac{6}{2}=3$.

Case $3$ is asking us how many ways there are to arrange $1$ $U$, which is $1$.

Case $4$ is asking us the same thing as Case $3$, giving us $1$.

Therefore, deducting all cases from $20$ gives $20-10-3-1-1=\boxed{\textbf{(B)}\,5}$.

~Tacos_are_yummy_1

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

https://youtu.be/BaE00H2SHQM?si=GTocuz7rsKFCrPn3&t=2986

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=n9jyl0QHXLbaKz3I&t=1363

~hsnacademy


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=-Yummy

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=1238

Video Solution by Dr. David

https://youtu.be/r3FtOOYEces

Video Solution by WhyMath

https://youtu.be/6Bg0Z0jcInw

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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