2005 Alabama ARML TST Problems/Problem 10

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Problem

When Jon Stewart walks up stairs he takes one or two steps at a time. His stepping sequence is not necessarily regular. He might step up one step, then two, then two again, then one, then one, and then two in order to climb up a total of 9 steps. In how many ways can Jon walk up a 14 step stairwell?

Solution

Solution 1

He can either take 14 one-steps, or 12 one-steps and 1 two-step, etc., so we have

\begin{eqnarray*} \frac{14!}{14!}+\frac{13!}{12!\cdot 1!}+\frac{12!}{10!\cdot 2!}+\frac{11!}{8!\cdot 3!}+\frac{10!}{6!\cdot 4!}+\frac{9!}{4!\cdot 5!}+\frac{8!}{2!\cdot 6!}+\frac{7!}{7!}\\ =1+13+66+165+210+126+28+1=14+231+336+29=245+365=\boxed{610} \end{eqnarray*}

Solution 2

Let $a_n$ represent the number of sequences of steps that can be taken up the stairs. Clearly $a_{1} = 1, a_{2} = 2$. Any sequence of length $n + 1$ is either composed of a sequence of length $n$ followed by another step or a sequence of length $n-1$ followed by two steps. Thus we have the recursion \[a_{n + 1} = a_{n} + a_{n - 1}\] which indeed is the Fibonacci sequence, except with indexes shifted. Matching them, we have $a_n = F_{n + 1}$, and $a_{14} = F_{15} = \boxed{610}$.

See also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 9
Followed by:
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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