2010 AMC 8 Problems/Problem 25

Problem

Everyday at school, Jo climbs a flight of $6$ stairs. Joe 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 enumerate all of the possibilities. We will denote a specific climb by using a sequence of $1s$ , $2s$ , and $3s$ in which each 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 two $2s$ 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.

2010 AMC 8 (Problems • Answer Key • Resources)
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2010 AMC 8 Problems/Problem 25

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