Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 8 Problems/Problem 25"

(→‎Solution)
(Added another solution.)
Line 20: Line 20:
  
 
Thus, there are <math>\boxed{\textbf{(E) } 24}</math> ways to get to step <math>6.</math>
 
Thus, there are <math>\boxed{\textbf{(E) } 24}</math> ways to get to step <math>6.</math>
 +
 +
==Solution 3==
 +
Complementary counting is also possible. Considering the six steps, Jo has to land on the last step, so there are <math>2^5=32</math> ways to land on the other five steps. After that, subtract the number of ways to climb the steps while taking a leap of <math>4</math>, <math>5</math>, or <math>6</math>. The eight possible ways for this is:
 +
(<math>4</math>, <math>1</math>, <math>1</math>)
 +
(<math>4</math>, <math>2</math>)
 +
(<math>1</math>, <math>4</math>, <math>1</math>)
 +
(<math>1</math>, <math>1</math>, <math>4</math>)
 +
(<math>2</math>, <math>4</math>)
 +
(<math>1</math>, <math>5</math>)
 +
(<math>5</math>, <math>1</math>)
 +
(<math>6</math>)
 +
 +
Altogether this makes for <math>32-8=24 \boxed{\textbf{(E) }</math> valid ways for Jo to get to step 6. ~adyj
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=24|after=Last Problem}}
 
{{AMC8 box|year=2010|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:51, 20 June 2020

Problem

Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$, $2$, or $3$ at a time. For example, Jo could climb $3$, then $1$, then $2$. In how many ways can Jo climb the stairs?

$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution 1

We will systematically consider all of the possibilities. A valid climb can be thought of as a sequence of some or all of the numbers $1$, $2$, and $3$, in which the sum of the sequence adds to $6$. Since there is only one way to create a sequence which contains all $1s$, all $2s$, or all $3s$, there are three possible sequences which only contain one number. If we attempt to create sequences which contain one $2$ and the rest $1s$, the sequence will contain one $2$ and four $1s$. We can place the $2$ in either the first, second, third, fourth, or fifth position, giving a total of five possibilities. If we attempt to create sequences which contain one $3$ and the rest $1s$, the sequence will contain one $3$ and three $1s$. We can place the $3$ in either the first, second, third, or fourth position, giving a total of four possibilities. For sequences which contain exactly two $2s$ and the rest $1s$, the sequence will contain two $2s$ and two $1s$. The two $2s$ could be next to each other, separated by one $1$ in between, or separated by two $1s$ in between. We can place the two $2s$ next to each other in three ways, separated by one $1$ in two ways, and separated by two $1s$ in only one way. This gives us a total of six ways to create a sequence which contains two $2s$ and two $1s$. Note that we cannot have a sequence of only $2s$ and $3s$ since the sum will either be $5$ or greater than $6$. We now only need to consider the case where we use all three numbers in the sequence. Since all three numbers add to $6$, the number of permutations of the three numbers is $3!=6$. Adding up the number of sequences above, we get: $3+5+4+6+6=24$. Thus, answer choice $\boxed{\textbf{(E)}\ 24}$ is correct.

Solution 2

A recursive approach is quick and easy. The number of ways to climb one stair is $1$. There are $2$ ways to climb two stairs: $1$,$1$ or $2$. For 3 stairs, there are $4$ ways: ($1$,$1$,$1$) ($1$,$2$) ($2$,$1$) ($3$)

For four stairs, consider what step they came from to land on the fourth stair. They could have hopped straight from the 1st, done a double from #2, or used a single step from #3. The ways to get to each of these steps are $1+2+4=7$ ways to get to step 4. The pattern can then be extended: $4$ steps: $1+2+4=7$ ways. $5$ steps: $2+4+7=13$ ways. $6$ steps: $4+7+13=24$ ways.

Thus, there are $\boxed{\textbf{(E) } 24}$ ways to get to step $6.$

Solution 3

Complementary counting is also possible. Considering the six steps, Jo has to land on the last step, so there are $2^5=32$ ways to land on the other five steps. After that, subtract the number of ways to climb the steps while taking a leap of $4$, $5$, or $6$. The eight possible ways for this is: ($4$, $1$, $1$) ($4$, $2$) ($1$, $4$, $1$) ($1$, $1$, $4$) ($2$, $4$) ($1$, $5$) ($5$, $1$) ($6$)

Altogether this makes for $32-8=24 \boxed{\textbf{(E) }$ (Error compiling LaTeX. Unknown error_msg) valid ways for Jo to get to step 6. ~adyj

See Also

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_25&oldid=126073"