# $n$ choose $k$ where $n$ is less than $k$

I am working on parameter estimation and one of the estimators involves a summation of $_nC_k$ ($n$ choose $k$) expressions. For some iterations, I need to compute expressions like $_0C_1$, $_0C_2$, etc.

In general how do we compute $_nC_k$ when $n$ is less than $k$? Do we still use the formula $\frac{n!}{(n-k)!k!}$ and use the gamma function to compute the negative factorial?

Thanks!

• By convention $\binom{n}k=0$ when $n<k$. – Brian M. Scott Dec 30 '12 at 14:16
• I take it $n$ is still a non-negative integer? – Gerry Myerson Dec 30 '12 at 14:20
• Note that you also can't "use the gamma function to compute negative factorial[s]" to get an answer, since the function is not defined on the non-positive integers. – TMM Dec 30 '12 at 14:26
• @TMM: Use the (entire) reciprocal gamma function instead, then: $$\binom nk = \frac{\frac{1}{\Gamma(n-k+1)} \frac{1}{\Gamma(k+1)}}{\frac{1}{\Gamma(n+1)}}$$ – hmakholm left over Monica Dec 30 '12 at 15:02

You could define $$_nC_k \text{ or } {n \choose k} = \frac{n(n-1)\cdots(n-k+2)(n-k+1)}{k(k-1)\cdots 2\cdot 1}$$ for positive integer $k$, and ${n \choose 0} = 1$ as the quotient of empty products.
This would give the usual values for non-negative integer $n \ge k$.
It would also give values for other real $n$. For non-negative integer $n \lt k$ including ${0 \choose k}$ it would give $0$ as the numerator involves multiplication by $0$ while the denominator does not.