What is $\binom{a}b$ with $a<b$?

Question

@Chris's_sis gave me following hint in a problem :

$\frac{1}{ \displaystyle \binom{ p+k}{p}}- \frac{1}{ \displaystyle\binom{p+k+1}{p}} =\frac{ p}{p+1}\frac{ 1}{\displaystyle\binom{p-k-1}{p-1}}$

However, in $\binom{p-k-1}{p-1}$ we have $p-k-1<p-1$

What is $\binom{a}b$ when $a<b$ ? Do we use $\Gamma$ and compute negative factorials ?

Answer 1

The value of ${a\choose b}$ is $0$ if $a<b$ and $a,b$ are both integers. There are two reasons for this, coming from two definitions of ${n\choose k}$

1. On one hand, ${a\choose b}$ is defined as "the number of $b$-sized subsets of a $a$-sized subset." Of course, a set with $a$ elements has no subsets of $b>a$ elements, so, by this definition, ${a\choose b} = 0.$

2. If you want to ignore the combinatorial definitions of ${a\choose b}$, you can simply define $${a\choose b} = \frac{a\cdot (a-1)\cdots (a-b+1)}{b!},$$ and, since $a-b+1 \leq 0$ and $a>0$, this means that one of the numbers $a,a-1,\dots, a-b+1$ must be zero, causing the whole product to equal zero.

Answer 2

Just expand the binomial coefficient: $\binom{\alpha}{k} = \alpha (\alpha-1) \ldots (\alpha -k +1)\cdot \frac{1}{k!}$. If $0< \alpha <k$, it's always 0.