Why is ${n-1 \choose -1} = 0$ when (-1)! is undefined? Shouldn't ${n-1 \choose -1}$ also be undefined?
$\endgroup$ 13 Answers
$\begingroup$By convention $\binom{n}k=0$ when $k>n$ or $k$ is a negative integer. This convention simplifies many manipulations involving sums and binomial coefficients. For example, we can write
$$\binom{n+1}k=\binom{n}k+\binom{n}{k-1}$$
without having to worry about whether $k=0$.
$\endgroup$ 8 $\begingroup$Here's one conceptual way to think of it:
$n\choose k$ is the number of ways to choose things $k$ things from $n$ objects. So $n-1\choose -1$ is the number of ways to choose $-1$ things from $n-1$ objects. Well, you can't choose $-1$ objects, so ${n-1\choose -1}=0$.
$\endgroup$ 1 $\begingroup$$\displaystyle\lim_{n\to\infty}n$ is undefined. $~\displaystyle\lim_{n\to\infty}\frac1n$ is $0.~$ The same happens here.
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