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I've come across a statement that says we have ${1} \choose {i-1}$$=0$ if and only if $i-1>1$. I don't know how to interpret the binomial coefficient when the bottom term is greater than the top. Why does this hold?

nomadicmathematician
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    From a set-theoretic point of view, the binomial coefficient $\begin{pmatrix} n \ k \end{pmatrix}$ gives the number of subsets of ${1,\ldots,n}$ of exactly $k$ elements. As no subset of ${1,\ldots,n}$ has more than $n$ elements, $\begin{pmatrix} n \ k \end{pmatrix} = 0$ for $k > n$ is justified. But of course, there might be different interpretations. – Maty Mangoo Aug 02 '23 at 14:20
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    The combinatorial definition of $\binom{n}{k}$ is the number of subsets of size $k$ of a set of size $n$. How many subsets of size $5$ has a set of size $3$? That's not the only way to define the binomial coefficient (all the definitions coincide when $n$ and $k$ are integers such that $0\le k\le n$), and it could happen that others definitions lead to different results when $k>n$. – jjagmath Aug 02 '23 at 14:20
  • [[A]] It might be better to change $i-1$ to $i$ , through out. It is unnecessarily complicated to use $i-1$ here. [[B]] More-over , the $1$ should be changed to some variable , to cover all Cases. – Prem Aug 02 '23 at 14:50
  • My guess would be tat you should be using $\binom{\alpha}k=\frac{\prod_{j=0}^{k-1} (\alpha-j)}{k!}$ as interpretation, but since here you seem to be discussing $\alpha\in\Bbb N$ other ways are possible. –  Aug 02 '23 at 14:50

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