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

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==Problem==
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== Problem ==
  
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BuzzY BUNNY BUNNY BUJFHGKDLA;SLDKFJ JFKDLS;ALSKDFJ ABCDEFGHIJKLMNOPQRSTUVWXYZ 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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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?
 
(For example, one sequence of hops is up-up-down-down-up-down.)
 
(For example, one sequence of hops is up-up-down-down-up-down.)
  
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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.

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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