Difference between revisions of "2024 AMC 8 Problems/Problem 13"
Scrabbler94 (talk | contribs) (Tag: Undo) |
m (Formatting) |
||
| (5 intermediate revisions by 4 users not shown) | |||
| Line 24: | Line 24: | ||
Counting all the paths listed above gets you <math>\boxed{\textbf{(B)} \ 5}</math>. | Counting all the paths listed above gets you <math>\boxed{\textbf{(B)} \ 5}</math>. | ||
| − | ~ | + | ~ALWAYSRIGHT11 |
| − | + | ~vockey(minor edits) | |
| + | ~Johnxyz1 (use mathit for better letter space) | ||
==Solution 2== | ==Solution 2== | ||
| Line 82: | Line 83: | ||
~Tacos_are_yummy_1 | ~Tacos_are_yummy_1 | ||
| − | ==Video Solution | + | ==Video Solution 1 by Central Valley Math Circle(Goes through the full thought process)== |
| + | https://youtu.be/aMmHkyEOmAc | ||
| + | |||
| + | ~mr_mathman | ||
| + | ==Video Solution 2 (don't understand the problem!)== | ||
https://youtu.be/BaE00H2SHQM?si=GTocuz7rsKFCrPn3&t=2986 | https://youtu.be/BaE00H2SHQM?si=GTocuz7rsKFCrPn3&t=2986 | ||
| − | |||
~Math-X | ~Math-X | ||
| Line 91: | Line 95: | ||
~hsnacademy | ~hsnacademy | ||
| − | |||
==Video Solution 3 by OmegaLearn.org== | ==Video Solution 3 by OmegaLearn.org== | ||
https://youtu.be/dM1wvr7mPQs | https://youtu.be/dM1wvr7mPQs | ||
| − | ==Video Solution | + | ==Video Solution 4(Easy to understand!)== |
https://www.youtube.com/watch?v=V-xN8Njd_Lc | https://www.youtube.com/watch?v=V-xN8Njd_Lc | ||
~NiuniuMaths | ~NiuniuMaths | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
Latest revision as of 16:41, 27 December 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5 (Complementary Counting)
- 7 Video Solution 1 by Central Valley Math Circle(Goes through the full thought process)
- 8 Video Solution 2 (don't understand the problem!)
- 9 Video Solution (A Clever Explanation You’ll Get Instantly)
- 10 Video Solution 3 by OmegaLearn.org
- 11 Video Solution 4(Easy to understand!)
- 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 
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 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 
.
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 1 by Central Valley Math Circle(Goes through the full thought process)
~mr_mathman
Video Solution 2 (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
Video Solution 4(Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=1238
Video Solution by Dr. David
Video Solution by WhyMath
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. 
