2024 AMC 8 Problems/Problem 13
Contents
- 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 (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 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 hops, and end up back on the ground? (For example, one sequence of hops is up-up-down-down-up-down.)
Solution 1
Looking at the answer choices, you see that you can list them out. Doing this gets you:
Counting all the paths listed above gets you .
~ALWAYSRIGHT11 ~vockey(minor edits) ~Johnxyz1 (use mathit for better letter space)
Solution 2
Any combination can be written as some re-arrangement of . Clearly we must end going down, and start going up, so we need the number of ways to insert 2 's and 2 's into . There are ways, but we have to remove the case , giving us .
- We know there are no more cases since there will be at least one before we have a (from the first ), at least two 'S before two 's (since we removed the one case), and at least three 's before three 's, as we end with the third .
~Sahan Wijetunga
Solution 3
These numbers are clearly the Catalan numbers. Since we have 6 steps, we need the third Catalan number, which is . ~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).
Therefore, we have .
vladimir.shelomovskii@gmail.com, vvsss
Solution 5 (Complementary Counting)
We can find the total cases then deduct the ones that don't work.
Let represent "Up" and represent "Down". We know that in order to land back at the bottom of the stairs, we must have an equal number of 's and 's, therefore six hops means of each.
The number of ways to arrange 's and 's is .
Case : Start with
Case : Start with
Case : Start with
Case : Start with
Case is asking us how many ways there are to arrange 's and 's, which is .
Case is asking us how many ways there are to arrange 's and , which is .
Case is asking us how many ways there are to arrange , which is .
Case is asking us the same thing as Case , giving us .
Therefore, deducting all cases from gives .
~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
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
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=1238
Video Solution by Dr. David
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.