# Binomial coefficient

1. Jul 8, 2015

### LagrangeEuler

1. The problem statement, all variables and given/known data
Calculate
$${-3 \choose 0}$$, $${-3 \choose 1}$$, $${-3 \choose 2}$$

2. Relevant equations
In case of integer $n$ and $k$
$${ n \choose k}=\frac{n!}{k!(n-k)!}=\frac{n(n-1)(n-2)...(n-k+1)}{k!}$$

3. The attempt at a solution
I am not sure how to calculate this. Any idea?

2. Jul 8, 2015

### haruspex

I've never worked with such things, but I believe binomial coefficients can be extended to noninteger and negative arguments by using the gamma function instead of factorials. Google it.

3. Jul 8, 2015

### LagrangeEuler

Yes but gamma function diverge for integer negative values. So $\Gamma(-3)=(-4)!=\infty$

4. Jul 8, 2015

### haruspex

Can you avoid that by some cancellation? You will have negative arguments in gamma functions both in the numerator and the denominator.

5. Jul 8, 2015

### hilbert2

The limit $\lim_{x\to n}\frac{\Gamma(x+1)} {\Gamma(k+1)\Gamma(x-k+1)}$ can exist even if $n$ is negative. Try it with Wolfram Alpha.

6. Jul 8, 2015

### LagrangeEuler

Not sure. For example ${-3 \choose 0}=\frac{(-3)!}{0!(-3-0)!}=1$. This is OK. But in case ${-3 \choose 1}=\frac{(-3)!}{(-4)!}=\frac{\Gamma(-2)}{\Gamma(-3)}=\frac{-3\Gamma(-3)}{\Gamma(-3)}=-3$
I think this is OK. Tnx :)

7. Jul 16, 2015

### Ray Vickson

The standard _definition_ of ${n \choose k}$ for non-negative integer $k$ and general real $n$ is given by the second formula you wrote in Post #1 (i.e., the formula that does not involve $n!$ or $(n-k)!$). Sometimes that formula can be expressed in terms of Gamma functions, and sometimes not.

Using that definition we obtain
$${-m \choose k} = (-1)^k {m+k-1 \choose k}$$
for integers $m,k \geq 0$.

Last edited: Jul 18, 2015
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