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

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==Problem==
 
==Problem==
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Everyday at school, Jo climbs a flight of <math>6</math> stairs. Joe can take the stairs <math>1</math>, <math>2</math>, or <math>3</math> at a time. For example, Jo could climb <math>3</math>, then <math>1</math>, then <math>2</math>. In how many ways can Jo climb the stairs?
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Everyday at school, Jo climbs a flight of <math>6</math> stairs. Jo can take the stairs <math>1</math>, <math>2</math>, or <math>3</math> at a time. For example, Jo could climb <math>3</math>, then <math>1</math>, then <math>2</math>. In how many ways can Jo climb the stairs?
  
 
<math> \textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24 </math>
 
<math> \textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24 </math>

Revision as of 10:25, 6 October 2015

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

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

An inductive 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 four 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.

See Also

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 809 		Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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