# How to evaluate binomial coefficients when $k=0$ and $1\geq|n|\geq0$

So I am doing some taylor series which boil down to the geometric series, for which I then need to evaluate various binomial coefficients. I have always used $n\choose k$ $=\frac{n!}{(n-k)!k!}$ to do these, this however becomes troublesome for non-integer and/or negative values of n.

Now I just memorized that if $k=0$, $n\choose k$ $= 1$ for integers $n\geq 0$. But this still leaves me with negative and/or negative values for n. My textbook provides me with $n\choose k$ = $\frac{n(n-1)...(n-k+1)}{k!}$, but I`m still not sure how to evaluate this with $k=0$. Somehow I always get it wrong.

For example i get $-1\choose 0$ $=\frac{-1(-1-0+1)}{0!}=\frac{-1*0}{1}=0$, but it should be $1$. Whats my error? Thanks!

I think you got confused by the notation $n(n-1)\dots(n-k+1)$. This is intended to mean the product of all the numbers you get when you start at $n$ and decrease, in steps of $1$ (to $n-1$ and so forth) until you reach $n-k+1$; so you don't include the factor that would come next, $n-k$. So there are $k$ factors here, the largest of which is $n$. Now when $k=0$, you're supposed to still use the same interpretation, not (as the notation made you reasonably think) start counting upward from $n$ instead of downward. So you should form a product of a sequence of numbers, starting at $n$ and decreasing in steps of $1$ down to and including $n+1$ (that's $n-k+1$) but not including $n$ (that's $n-k$). So you're supposed to stop before you start! In other words, when $k=0$, the expression $n(n-1)\dots(n-k+1)$ is intended to mean the product of no factors, which is $1$, not the product of two factors as you supposed.

• So this is the error I make, thank you for pointing it out to me. So for any value of n, if k=0 I always take the product of no factors, which is independent of n.
– Leo
Jun 8, 2013 at 21:47

For $\binom{-1}{0}$ take $\binom{\alpha}{0}=\frac{\alpha!}{(\alpha-0)!0!}=\frac{1}{1 \cdot 0!}=1$. Along these lines, e.g. $\binom{-1}{10}=\frac{\alpha}{10}=\frac{1}{10!}=\cdots (-1)^{10}\frac{10!}{10!}=1, \ \binom{-1}{11}=(-1)^{11}=-1$, etc

• Thank you. One thing I do not understand. How do you evaluate (-1)! ? I use the definition without $n!$ to avoid taking factorials of negative and fractured numbers.
– Leo
Jun 8, 2013 at 21:20
• You don't: it does not arise in these calculations. $(-1)!=\Gamma(0)$, which is some form of infinity, but I'm not sure which.
– Alex
Jun 8, 2013 at 21:22
• I shouldnt read $(-1)!=1$ from $\frac{\alpha!}{(\alpha-0)!0!}=\frac{1}{1 \cdot 0!}$ if $\alpha=-1$? I only know to do factorials by multiplying the input with itself minus one until i reach one. I dont know the $\Gamma(\alpha)$ function.
– Leo
Jun 8, 2013 at 21:29
• No in this case $\alpha!$ cancel out, so you have just $1$ left
– Alex
Jun 8, 2013 at 21:31
• So if $k=0$ you are always left with $\frac{\alpha!}{\alpha!}$ which is always 1, even for negative and fractured values of $\alpha$?
– Leo
Jun 8, 2013 at 21:41