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Find the number of ways to select $k$ objects from available $(n-k)$ objects, for all possible values of k such that $k < n/2$ and assume $n$ to be odd.

What I have inferred from this till now is that I need to find the sum of the series

$$\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+ \ldots + \binom{\frac{n+1}{2}}{\frac{n-1}{2}}$$

Please help me understand, how to convert it to some formula in $n$. Somewhere I saw it had some equivalence with $Fib(n)$ (i.e., $n^{th}$ term in Fibonacci sequence). I couldn't derive the intuition behind it either.

cosmo5
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abhi gang
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1 Answers1

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The expression counts the number of ways a set of $n$ stairs can be climbed by taking steps one or two at a time. If exactly $k$ moves involve taking two steps at a time, then stairs are climbed one at a time $n - 2k$ times, so a total of $n - 2k + k = n - k$ moves must be made. Hence, the total number of ways of climbing a set of $n$ stairs in which each move consists of moving one either one step or two steps at a time is $$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n - k}{k}$$ As you should verify, the $n$th term of this sequence is $F_{n + 1}$, where $F_n$ is the $n$th Fibonacci number

N. F. Taussig
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  • This surely game me some better insight , but I'm still confused how can I verify this equivalence , cause the other proofs mentioned are quite complex too . – abhi gang Oct 18 '20 at 11:38
  • thanks for trying , but the link to question mentioned by @robjohn was what i was looking for . – abhi gang Oct 18 '20 at 19:04