In how many ways can one climb a staircase composed of 10 stairs when each step covers either one stair or two stairs?
This question is very similar to one I have seen with 6 stairs. The solution was to find how many possibilities are there to take one stair at a time, then 2, then 3, and so on. I need help because the number is larger.
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The number of $\ds{n}$-steps configurations is given by
\begin{align} &\sum_{s_{1} = 1}^{2}\ldots\sum_{s_{n} = 1}^{2} \bracks{s_{1} + \cdots + s_{n} = 10} = \sum_{s_{1} = 1}^{2}\ldots\sum_{s_{n} = 1}^{2} \oint_{\verts{z} = 1}{1 \over z^{11 - s_{1} - \cdots - s_{n}}} \,{\dd z \over 2\pi\ic} \\[5mm] = &\ \oint_{\verts{z} = 1}{1 \over z^{11}}\pars{\sum_{s = 1}^{2}z^{s}}^{n} \,{\dd z \over 2\pi\ic} = \oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{11 - n}}\,{\dd z \over 2\pi\ic} = {n \choose 10 - n} \end{align} which is non zero whenever $\ds{\pars{~n \geq 10 - n \geq 0 \implies 5 \leq n \leq 10~}}$.
The total number of configurations is given by
$$ \sum_{n = 5}^{10}{n \choose 10 - n} = \sum_{n = 0}^{5}{n + 5 \choose 5 - n} = \bbx{\ds{89}} $$
