Difference between revisions of "2024 AMC 8 Problems/Problem 13"
Niuniumaths (talk | contribs) (→Video Solution by CosineMethod [🔥Fast and Easy🔥]) |
(→Solution 3) |
||
| Line 32: | Line 32: | ||
These numbers are clearly the Catalan numbers. Since we have 6 steps, we need the third Catalan number, which is <math>\boxed{\textbf{(B)}\ 5}</math>. | These numbers are clearly the Catalan numbers. Since we have 6 steps, we need the third Catalan number, which is <math>\boxed{\textbf{(B)}\ 5}</math>. | ||
~andliu766 | ~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 <math>1 + 2 \cdot 2 = \boxed{\textbf{(B)}\ 5}</math>. | ||
| + | |||
| + | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Video Solution 1 (easy to digest) by Power Solve== | ==Video Solution 1 (easy to digest) by Power Solve== | ||
Revision as of 06:43, 27 January 2024
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.)
Contents
- 1 Solution 1
- 2 Solution 2
- 3 Solution 3
- 4 Solution 4
- 5 Video Solution 1 (easy to digest) by Power Solve
- 6 Video Solution 2 by Math-X (First fully understand the problem!!!)
- 7 Video Solution 3 by OmegaLearn.org
- 8 Video Solution by NiuniuMaths (Easy to understand!)
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 See Also
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 5 or B.
-ALWAYSRIGHT11
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).
Totally we have 
.
vladimir.shelomovskii@gmail.com, vvsss
Video Solution 1 (easy to digest) by Power Solve
Video Solution 2 by Math-X (First fully understand the problem!!!)
https://www.youtube.com/watch?v=Td6Z68YCuQw
~Math-X
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
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. 
