2024 AMC 8 Problems/Problem 13

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5 (Complementary Counting)
7 Video Solution 2 by Math-X (don't understand the problem!!!)
8 Video Solution (A Clever Explanation You’ll Get Instantly)
9 Video Solution 3 by OmegaLearn.org
10 Video Solution by NiuniuMaths (Easy to understand!)
11 Video Solution by CosineMethod [🔥Fast and Easy🔥]
12 Video Solution by Interstigation
13 Video Solution by Dr. David
14 Video Solution by WhyMath
15 See Also

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.)

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

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 (don't 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

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

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