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 (First fully understand the problem!!!)
8 Video Solution 3 by OmegaLearn.org
9 Video Solution by NiuniuMaths (Easy to understand!)
10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
11 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:

$UUDDUD$

$UDUDUD$

$UUUDDD$

$UDUUDD$

$UUDUDD$

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

~ALWAYSRIGHT11 ~vockey(minor edits)

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

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 $D$

Case $2$ : Start with $UDD$

Case $3$ : Start with $UUDDD$

Case $4$ : Start with $UDUDD$

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 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

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
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